TK 2625 
. A5 

Copy 1 


DESIGN OF 


Direct Current Dynamos. 

American 

School of Correspondence 

Copyrighted 1898 

BY 

American School of Correspondence 


BOSTON, MASS., 
U. S. A. 





fT9f" -**« yrt'J-/ 


DESIGN OF 

Direct Current Dynamos. 


INSTRUCTION PAPER 

/ 

American 

f \ 

School of Correspondence 


Copyrighted 1898 

BY 

American School of Correspondence 


BOSTON, MASS., 
U. S. A. 









7/f itiS" 

, /l 6 













\ 


18280 ^ 























INTRODUCTION, 


Having become familiar with some of the fundamental principles 
of the dynamo by a study of electromagnetic induction in the 
Instruction Paper on Theory of Dynamo-Electric Machinery , one 
is in a position to gain a more thorough understanding of the dynamo 
by making a more minute study of the details of actual machines. At 
first some of the principal types of direct current dynamos will be dis¬ 
cussed, and then in order to bring details clearly before the mind the 
various parts of a complete dyuamo will be designed and the principles 
involved will be explained. The method of calculating used will be 
that adopted by Mr. A. E. Wiener, and for a more thorough study of 
this subject the student is referred to his book on Dynamo-Electric 
Machines. Symbols recommended by the Electrical Congress will be 
used together with a few others that have come into use verv lately. 
A complete list of the notation as used throughout the Instruction 
'Paper is also collected for easy reference. The development of this 
subject will also require a use of some of the absolute units which will 
be discussed briefly on one of the following pages. 

In the design of a dynamo-there must be considered, first the 
proper relation of each part of the dynamo to the whole, and second 
the relation of the electrical and magnetic circuits. No attempt can 
be made in this general course to give details of design for all types 
and all sizes of machines. Mr. Wiener’s Practical Calculation of 
Dynamo-Electric Machines treats this subject very fully and very 
clearly, and the formulas, tables and calculations in his book are 
arranged for both the English and the Metric systems of measurement. 

The Shunt, Series, and Compound Dynamos have already been 
treated briefly in the Instruction Paper on theory of Dynamo-Electric 
Machinery. They will be still further discussed in the present paper, 
and their various characteristics will be treated mathematically, and 
again graphically by means of curves. Much of the mathematical treat¬ 
ment has been made Optional. Those wishing only a general knowl¬ 
edge of the subject may omit all sections so marked. Those 
wishing a more exact mathematical knowledge of the subject will do 
well to study all the sections. The mathematics will be found very 
simple, comprising only the applications of Ohm’s law, simple arithmetic, 
and very elementary algebra. In discussing the design of a dynamo, 
first the principles are treated, and then the equations for dimensions 
are given. All the equations may be considered as Optional although 
the descriptive matter concerning the principles is very important and 
should be studied very carefully. 



SYMBOLS FOR PHYSICAL QUANTITIES 


Recommended by the Committee on Notation of the Chamber of 
Delegates of the International Electrical Congress of 1893. 

With the names added in italics of the practical magnetic units provisionally 
adopted by the American Institute of Electrical Engineers. 


Physical Quantities. 

Symbols. 

Fundamental. 



A, 1 


M 


T , t 

Geometric- 


S , S 


V 

Angle. 

a , ft 

Mechanical. 


Velocity....-.-. 

V 

Angular Velocity. 

CO 

Acceleration.. 

a 


E,f 

Work.-.. 

w 

Power. .. 

p 

Pressure. ... 

P 

Moment of Inertia. 

K 

Magnetic. 


Strength of Pole ... 

m j 

Magnetic Moment..... 

on 

Intensity of Magnetization. 

0 

Field Intensity. 

x 

Flux of (Magnetij:) Force.. 


Magnetic Induction. 

($> 

Magnetizing ForceJ... .. 

X 

Magnetomotive Force. 

<r 

Reluctance (Magnetic Re- 

01 

, sistance). 


(Magnetic) Permeability. ... 

p 

(Magnetic) Susceptibility... 

x 

Reluctivity (Specific Mag- 


netic Resistance). ......... 

V 

Electromagnetic. 


Resistance. 

R . r 

Electromotive Force. 

E . e 

Difference of Potential.... 

U , u 

Intensity of Current. 

J , i 

Quantity of Electricity. ...., 

Q . <i 


C , e 

IV 

Electric Energy. 

Electric Power. 

p 

Resistivity (Specific Resist- 


ance)... 


Conductance. 

d, g 

Conductivity (Specific Con- 


ductance). 

r 

Coefficient of Induction (In- 


ductance).... 



[ 



Dimensions of 

Equations.' 

the Physical 
Quantities. 


E 


M 

T 


S = L.L 

i ;* 

V-L.L.L 

z 3 

arc 

A number. 

radius ' 
L 


V ~ ~T 

FT-' 

V 



T- x 

V 


a =T 

LT -* 

F- Ma. 

LMT-' 

IV = PL 

L‘M T- t 

„ w 


P ~~T 

r-MT-* 

F 


^ = V 

L-'AIT-‘ 

v K = MV 

JAM 

li 

*1* 

IPIY T-' 

on. = »‘i 

jfaYt-' 

« 97L 

E^aYt-' 

X = - 

L~ i M i 7-' 

1)1 


■ $ = XV 

Eh/ip- 1 

ffi = /'X 

E~ i Ai i r- 1 

TP _ ATTNl 

L 

L^AY 7- 1 

$7 = 4 tcNI 


Ot -yL 

Z J 

s 


© 


P “ X 

A number. 

3 

K — - 

A number. 

X 


i 


V =r — 

A number. 

M 


£ 


R ~1 

LT- 1 

E - RI 

jaaY t-' 

U= R1 

E'A'Yt-'* 

r-f 

E*M*T- 1 

Q — IT 

E i A/i 

C-% 

z- 1 r 

£ 


IV = EIT 

JL'MT -* 

P= El 

LA All-' 

RS 

Z* T- 1 

P = ~L 

Cl 

II 

L-'T 

,R 


1 

y = - 

L-'T 

p 


$ 


L-~ 

L 

i 



Names of the C. G. S. 
Units. 

Practical Units. 

Centimetre. 

Mass of one gramme. 
Second. 

Metre. 

Mass of a kilogramme.' 
Minute ; hour. 

Square centimetre. 
Cubic centimetre. 

Square metre. 

Cubic metre. 

Radian. 

Degree ; minute ; 
second ; grade. 

Centimetre per second. 

Metre per second. 

Radian per second. 

Revolutions (turns) per 
minute. 

Centimetre per second 
per second. 

Dyne. 

Erg. 

Metre per second per 
second. 

Gramme; kilogramme. 
Kilogrammetre. 

Erg per second. 

Kilogrammetre per 
second. 

Dyne per square centi¬ 
metre. 

Gramme-mass-centime¬ 

tre-squared. 

Kilpgram per square 
centimetre. 

Gauss. 

Gauss. 

Weber. 

Gauss. 

Weber. 

Gauss. 

Gauss. 

Gauss. 

Gilbert. 

Oersted. 

Gilbert; I Gilbert^ 0.7958 
ampere-turns (a-t). 
Oersted. 


Ohm. 

Volt. 

Volt. 

Ampere. 

Coulomb; ampere-hour 
Farad. 

Joule ; watt-hour. 
Watt; kilowatt. 

Ohm-centimetre. 

Mho. 


Henry. 


t iV Is the number of windings, and L the length of the solenoid 
ing force. 


generating the magnetiz- 

































































DESIGN OF 


DIRECT CURRENT DYNAMOS. 


In the study of Direct Current Dynamos it will be of advan¬ 
tage to divide the subject into two parts ; (1) Constant Potential 
Dynamos, machines that supply current at a constant pressure for 
all loads ; and (2) Constant Current Dynamos, machines that 
supply a current of constant strength for all loads. 

It has been explained how the energy in a circuit is equal to 
the product of the difference of potential, E , between the terminals 
of the dynamo, and the current, J, flowing in the circuit, or E I. 
When this product E 1 is large, a large amount of work is expended 
in the circuit, and to express this fact it is said that a heavy load 
has been put upon the dynamo. The value of E I is increased by 
increasing either E or 7, or both, that is, the load on the dynamo 
is increased when a larger current comes from it, or when the same 
current flows but at a higher voltage. 

By having more lamps or running more motors on the circuit 
we increase El, or the load upon the dynamo. In constant poten¬ 
tial machines, E remains constant for any change in the number 
of lamps or motors on the circuit, within the limits of the capacity 
of the machine, hence to increase the load on the dynamo, /must 
be increased, and it is seen readily enough that the load is pro¬ 
portional to I if E remains constant, or since I is inversely propor¬ 
tional to R, the resistance of the circuit, the load on a constant 
potential machine is inversely proportional to the resistance of the 
circuit. 

In a constant current machine I is constant and E must be 
increased if E I, the load, is to increase. If E rises, and I remains 
the same, then R must increase according to Ohm’s law. Hence 
the load on a constant current machine is proportional to E and 
to R. An increased load on constant current machines, requires 
larger values of E and R ; on constant potential machines it 
requires a larger value of I and a smaller value of R. 



6 


DIRECT CURRENT DYNAMOS. 


If E is to remain constant the magnetization produced by the 
field coils must remain constant, or in other words the conductors 
on the armature must cut the same number of magnetic lines of 
force each second. If I is to remain constant and E is to increase 
then a correspondingly greater number of lines of force must be 
cut each second. 

The essential difference between the two types of machines 
then is that in one there is a nearly constant magnetic flux for all 
loads, and in the other there is a magnetic flux varying with the 
load. The two types of constant potential and constant current 
dynamos will be considered separately. 

Constant Potential Dynamos. 

For supplying incandescent lamps in parallel, or for running 
motors to furnish power at a constant speed, a constant electro¬ 
motive force is required at the terminals of the lamps or at the 
brushes of the motor. As the number of lamps in multiple is 
increased, or as more load is thrown on the motors, or as more 
motors are started, there are more paths prepared for the current 
in the outside circuit, and more current flows from the dynamo. 
The voltage remains constant, and the resistance of the circuit 
being less, the current becomes greater. 

It has been pointed out, however, (Inst. Paper on Theory of 
Dynamo-Electric Machinery , pp. 35-36) that on account of the 
armature resistance the lost voltage in the armature increases with 
the load, and in addition to this an increase of load increases the 
armature cross magnetizing effect which tends to weaken the field 
and therefore to still further decrease the voltage. Therefore if 
the speed of the shunt dynamo is not altered, an increase of load 
will decrease the effective voltage by the amount of increase in the 
lost voltage, and by the amount of decrease in the field strength. 
The result will be to make the lamps burn less brightly for every 
extra one put on the circuit, or in the case of a power circuit the 
motors would slow down a few revolutions each time the load on 
the dynamo was increased. 

As the electromotive force of a dynamo is directly propor¬ 
tional to the speed, to the number of armature conductors, and to 
the magnetic flux due to the field coils, so it may be changed in 




DIRECT CURRENT DYNAMOS. 


7 


value by changing the value of any one of these factors. Although 
it is possible to change the effective value of the number of con¬ 
ductors on the armature by means of a change in the position of 
the brushes, and although it is possible to change the speed of the 
machine by means of gearing, yet neither of these methods are 
convenient, nor do they prove very satisfactory. It is possible, 
however, to change the value of the magnetic flux produced by 
the field coils quite easily and to regulate the value of this change 
quite accurately. 

Suppose that the number of ampere-turns needed for a given 



machine has been determined (the method for doing this will be 
explained later) and suppose that it has been decided what pro¬ 
portion of the ampere-turns shall be in amperes and what propor- 
























































8 


DIRECT CURRENT DYNAMOS. 


tion in turns so that the size of the wire is fixed. Then if a 
somewhat larger wire than the one required be used the resistance 
of the field coils will be less and the field current larger than it 
was figured. By inserting a rheostat, or resistance which can be 
varied at will, between the field coil and one of the brushes, the. 
current in the coil can be cut down to the proper value under 
normal conditions, and when the load increases and the effective 
voltage tends to drop, it is only necessary to cut out some of the 
resistance in the rheostat to obtain more current, and consequently 
greater excitation in the field coil, thereby raising the voltage to its 
normal value again. The rheostat (see Fig. 1) consists generally 
of a resistance with connections at short intervals along it and 
terminating in metal contacts arranged in a circle. A sliding con¬ 
tact piece is moved over them by the handle, introducing the cur¬ 
rent at any point on which the sliding contact is resting. In this 
way the amount of resistance in series with the field coil may be 
regulated with great exactness. 

This regulation is generally effected by hand, the resistance be¬ 
ing decreased when the voltmeter shows that the voltage is falling 
on the line and vice versa. By means of regulators, magnets, solen¬ 
oids and various electrical and mechanical means this regulation 
is sometimes made automatic. 

Efficiency of Dynamos. 

No machine has ever yet been built so perfect that it is able 
to apply to the work for which it was built all the energy which 
it receives. Some of it is used up in keeping the machine itself 
running ; this is called waste energy, and goes to overcome friction 
of some kind. The more efficient the machine the less is the waste 
energy. This waste is undesirable for two reasons, first because 
it deducts something from the output of the machine, and second 
because though wasted it is not lost, and reappears as heat. Every 
one is familiar with the effects of a hot box. The energy going to 
waste to overcome friction in the journal is converted into heat 
and raises the temperature of the parts. If the heat is not car¬ 
ried off by oil the parts may enlarge slightly which in turn causes 
more friction, more and more power is absorbed, and the machine 
rendered more and more inefficient until it may even stop entirely, 



DIRECT CURRENT DYNAMOS. 


9 


the energy supply being insufficient to overcome the friction; the 
efficiency is then zero. 

Efficiency is usually expressed in terms of the energy delivered 
to the machine and is given as the relative percentage that the out¬ 
put is of the input. Modern commercial dynamo machines of fair 
size reach an efficiency as high as 92 %, that is they deliver at their 
brushes electrical energy equal to of the mechanical energy 
furnished them at the pulley. 

Part of the wasted energy is used in overcoming mechanical 
friction in bearings and brushes, most of this being in the bear¬ 
ings. This should not exceed 5% and in many cases is much less, 
depending on the size, design and purpose of the generator. The 
remainder of the energy is accounted for in the electrical losses, 
for although, as previously explained, the energy is not lost 
since it reappears as heat, it is of course lost from the practical 
standpoint. The resistance of wire to the passage of an electrical 
current is analogous to mechanical friction. It requires power to 
force a current through a wire although it need only be a very 
small portion of the total power that can be transmitted by the 
current. The evidence of the waste of power appears in the loss 
of voltage which takes place between the beginning and end of the 
wire and this power reappears as heat in the wire. Evidently 
therefore there must be a loss in the windings of the dynamo due 
to the current circulating in them, and this loss is usually con¬ 
sidered under the heads of armature loss and field loss. By mak¬ 
ing the wire large and therefore reducing the resistance, this loss 
may be reduced to as small a value as desired, the limit being found 
in expense and practicability of construction. 

There is another electrical loss termed core loss which results 
from that property of iron which has already been dealt with under 
the name of hysteresis. This is also a kind of friction and the 
power expended in overcoming it reappears as heat in the iron 
core. (See page 25 Instruction Paper on Theory of Dynamo-Electric 
Machinery.) Hysteresis is increased by hardness in the metal and 
increase of rapidity in the changes of magnetism to which it is 
subjected. For this reason the softest possible iron or steel is 
selected for armatures since they are subjected to one reversal 
of magnetism for every revolution they make. There is a very 



10 


DIRECT CURRENT DYNAMOS. 


small additional loss due to eddy currents. This is the result of 
setting up currents which are not useful or desirable in metallic 
parts of the machine. This effect is very successfully avoided by 
laminating the armature core, which consists in building it up of 
thin sheets of iron whose planes are at right angles to the conduc¬ 
tors. The faces of the laminations are coated with varnish, or have 
paper between them, or are allowed to rust, so as to be coated 
with an iron oxide, which effectually stops any currents which 
tend to generate in the iron mass parallel to the conductors. The 
core losses and eddy current losses together should not exceed 2 %. 

The shunt=wound dynamo has high resistance field coils or 
many turns of fine wire. As the terminals of the field coils as 
well as the terminals of the outside circuit are both connected to 
the brushes there are two paths through which the current may 



Fig. 2. Diagram of Shunt-Wound Dynamo, 
flow (see Fig. 2). The path leading to the outside circuit where 
the commercial results are accomplished is considered the circuit 
of first importance and the path through the field coils, which 
however is important to the proper working of the dynamo, is 
termed a shunt around the other path. 

The important formulas showing the characteristics of the 
shunt dynamo will be developed. 

Let I' = total current in armature; 

I sh current flowing in shunt coils; 

I = useful current flowing in outside circuit; 

E' = total E. M. F. generated in armature; 

E = potential of dynamo, both for shunt coil, and out¬ 
side circuit; 

r a = resistance of armature ; 

r sh = resistance of shunt field coils; 

R = resistance of outside or useful circuit; 

rj e = electrical efficiency. 









DIRECT CURRENT DYNAMOS. 


11 


These various quantities can be expressed in terms of each 
other in a number of different ways. This is of value in studying 
and understanding the characteristics of the shunt dynamo, and in 
finding the values of unknown terms when others are known. It 
is obvious by looking at diagram of Fig. 2, that the total current 
is the sum of the outside current and shunt current, or 

T = I+I sb or = 

^*sh 

This latter equation shows what amount of the armature cur¬ 
rent is lost in the shunt field. Also the total current in the arma¬ 
ture is equal to the total E. M. F. divided by the total resistance. 


r = 


E' 




i R X r sh 

R + '•sh 


for the joint resistance of the shunt and outside circuits (See 
Instruction Paper on the Electric Current , p. 17) is equal to the 
product of the two divided by the sum of the two. 

*The relation can be expressed in still another way since 


^ = I and I = from which I’ == j 1 + \ 

R -^sh R R \ r gb ) 

E f R , and r sh are often known quantities from which I 1 and 
E r may be found by the aid of these relations. The value of the 
useful current in the outside circuit is 



and the value of the current in the shunt current is 


f The value of the total E. M. F. generated by the dynamo is 
E’ = E + r r a 

from which it may be seen by inspection that the lost volts in 
the armature E f — E, increase directly with I 1 . If E’ and r a are 
considered as remaining constant, then an increase in 1’ increases 
iV a also, and E must become less. If E is decreased and E' is 


* Optional. 


-j- Study. 










12 


DIRECT CURRENT DYNAMOS. 


constant then their difference which is equal to the lost volts in 
the armature must increase. 


*As I' = I ^shi we have by multiplying by r a and substi¬ 
tuting for j TV a its value E' — A', 

E< = U+ Ir & + J sh r a = Z+ 4 ra + A r a 

»'sh 


= jr 


1 + 

-R ^ sh 


= EXr e 


A + A + A 

J! r ah 


f The electrical efficiency of the dynamo is equal to the value 
of the useful energy divided by the total energy developed in the 
armature, the total energy itself being equal to the useful energy 
plus the energy lost in the armature, plus the energy lost in the 
field coils. Therefore the electrical efficiency is 

El EE 


Ve 


ET P B + + P ah r Bh 


*As I* = I X ^ rsh ? and 7 sh = '^L we obtain 

^sh r ah 


Ve — 


P E 


PE+ \ I X 


E 


^*sh 


^a + 


IE 


^*sh 


Ve =- 

PE + 


72 R 


i p w ± 2 PJl + P r 3 sh 



P R 2 

^*sh 


1 ft ^'a I ! (7a 

r sh ^ A ^ 

By a mathematical analysis it can be proven that the value of r) e is 
a maximum when 


-R = r s h y 


r z + »\* 


This equation shows what the resistance of the outside circuit of a 
dynamo should be, knowing the shunt and armature resistances, 
in order to have the machine carry the proper load for maximum 
electrical efficiency. 















DIRECT CURRENT DYNAMOS. 


13 


* Replacing R by its value in terms of r a and r sh we have the 
terms 


R Osh + >~a) = r Bh + W J r a _ sh\ (r sh + r a ) 
^ 2 sh r A V j- a +r sh r sh 


and r J^ = l l a t Aa + i’ S h _ vV»(>. + r. h ) - 

-B '^h V r sh 

t The equation of the maximum efficiency of a shunt dynamo 
becomes 


^e.Max. — 


1 

1 _|_ 2 C^a, ~j~ ^sh) | 2 r a 

^*sh 7*sh 


The value of the armature resistance compared with the value 
of the shunt resistance is always so small that r a -f- r sh may be 

written r sh without any appreciable error and the value of r& — is 

^sh 

so small that it may be neglected, then the approximate value of 
the electrical efficiency of the shunt dynamo may be written in 
very simple terms and may be considered as equal to 



and the approximate ratio of the shunt resistance to the armature 
resistance is expressed by the formula 

fsb. = ( 2 Ve j 2 

r ^ ^ 1 * Ve 1 


The following table from Wiener gives the necessary ratios 
of the resistance of the shunt field to the resistance of the arma¬ 
ture for various electrical efficiencies for shunt dynamos. As these 
values are not given in ohms but only as ratios they hold good for 
all shunt dynamos, large or small. It will be noted that this 
electrical efficiency may be as near 100% as is desired, from a 
theoretical standpoint, but the limit is generally governed by the 
practical consideration of the cost. If the armature resistance is 
made too small, more copper is used to carry the current than is 
required, and if the field coil resistance is made extremely high the 
wire will be very small and very expensive. 


















14 


DIRECT CURRENT DYNAMOS. 


TABLE GIVING RATIO OF SHUNT TO ARMATURE RESISTANCE FOR VARIOUS 
ELECTRICAL EFFICIENCIES FOR SHUNT DYNAnOS. 


Electrical 

Efficiency. 

Ratio of 
Shunt to 
Armature 
Resistance. 

Electrical 

Efficiency. 

Ratio of 
Shunt to 
Armature 
Resistance. 

100 X Ve 

£sh 

r a 

100 X Ve 

Tih 

U 

80f 0 

64 

96 

2,304 

85 

128 

97 

4,182 

90 

324 

98 

9,604 

93 

706 

99 

39,204 

95 

1,444 

99.5 

158,404 


A number of very important characteristics of a dynamo 
obtained from experiment may be shown by means of what are 
called Characteristic Curves. These same characteristics may 
generally be figured out mathematically but it is more correct and 
more satisfactory to get at the results from a practical test as a 
general rule. These characteristic curves explain the actions and 
possibilities of a dynamo in very much the same way that the 
indicator card of a steam engine gives its desired information. By 
drawing the characteristic curves to some known scale the horse¬ 
power at which the dynamo will work with the greatest efficiency 
can be calculated. 

A Saturation Curve shows to what degree the iron has been 
saturated with magnetism and thus whether or not it is properly 
designed and proportioned as regards its magnetic circuit. Fig. 3 
shows such a curve. The curves are formed by plotting points 
whose abscissae are values of ampere-turns and whose ordinates 
are the corresponding values of the voltage, these points being plot¬ 
ted for varying values of the current. The readings may be 
observed for both ascending and descending values of the current 
thus showing the effects of hysteresis. By means of two electro¬ 
magnets, one being placed in series in the circuit so as to pull in 
proportion to the current, and the other being a shunt so as to pull 
with a force proportional to the voltage, it is possible to cause 
a dynamo to draw its own characteristic curve just as a steam 
engine draws its own indicator card. 


















DIRECT CURRENT DYNAMOS. 


15 



Fig. 3. 



















































































































































16 


DIRECT CURRENT DYNAMOS. 


Having drawn the characteristic curve of a machine to 
some particular scale, and knowing either the current, /, or volt¬ 
age, A 7 , at which the dynamo is working, the other value may be 
read off the curve directly. The value of the current multiplied 
by the corresponding voltage gives the output of the machine in 
watts and this divided by 746, the number of watts corresponding 
to one horse-power, gives the output of the machine in mechani¬ 
cal units or in horse-power. 

It is possible to plot horse-power curves on the same paper 
with the characteristic curve (see Fig. 4) since horse-power = 
volts X^am peies^ Any combination of amperes X volts which 

will give 746 is equal to one horse-power, thus 7.46 amp. X 10 
volts = 1 horse-power. Evidently then there is an infinite 
number of points the product of whose co-ordinates equals 746, 
and a line drawn through these points is a curve of one horse¬ 
power. If the characteristic of the dynamo cuts the curve at any 
point it means that at that point the dynamo is furnishing one 
horse-power of electrical energy. It follows at once that curves 
may be similarly drawn for 2, 3, 4, or any number of horse-power. 
Then a glance at the characteristic curve will reveal at once 
what the activity of the dynamo is. 

The dynamo has two characteristic curves, the External Char¬ 
acteristic found by plotting the voltage and the current of the 
outside circuit and the Internal Characteristic found by plotting 
the total voltage and the total current generated. The character¬ 
istic due to series coils is similar to a curve of magnetic induction, 
the value of the voltage increasing rapidly at first with the increase 
of the ampere-turns; then as the iron core becomes more or less 
saturated with magnetic lines the voltage increases more slowly 
as the ampere-turns or current increase. The shunt characteristic, 
however, has a maximum value of voltage (see Fig. 4) when the 
external current is zero and falls as the current increases, due 
partly to the demagnetizing effects of the armature and partly on 
account of the ever increasing lost volts in the armature which 
increase with the current. At first the voltage falls as the current 
or load increases but finally the capacity of the dynamo is reached 
and after a certain fixed point both voltage and current decrease 
and fall to zero very rapidly, as shown in the figure. 





DIRECT CURRENT DYNAMOS. 


IT 



Fig. 4. 


























































































































































































































































































18 


DIRECT CURRENT DYNAMOS. 


The resistance of the outside circuit corresponding to any 
point on the characteristic curve can be found very readily. The 
voltage divided by the current gives the resistance. By drawing 
a straight line (see Fig. 4) from the origin or the point where the 
current and voltage are 0 through the point corresponding to 20 
volts and 10 amperes it will intersect the characteristic curve at 
a certain point. It is obvious from the figure that the voltage at 
this point is just double the current and so the resistance here is 
2 ohms. Similarly by drawing a line from the origin through the 
point where voltage is 40 and current 10 the characteristic curve 
is intersected at another point and the resistance of the circuit at 
this point is 4 ohms. A scale of resistance may be marked on 
the vertical line corresponding to 10 amperes. By drawing 
a line from any point on the characteristic curve to 0, 
the point of intersection on this resistance scale line gives the 
resistance of the outside circuit. The characteristic curve shown 
in Fig. 4 is for a small shunt dynamo which being run at G30 
revolutions gave a maximum of a little less than 2 horse-power at 
4T.5 volts and 30 amperes. With a decrease in resistance much 
below 2 ohms the current increases but little, whereas the voltage 
falls a great deal. If the external resistance becomes less than 1 
ohm the machine loses its voltage and current immediately and 
will not build itself up again until the resistance is increased. 
The most critical part of this curve is where the voltage is about 
30 or 31 and any given change of resistance at this point will alter 
the voltage more than at any other part of the curve. By increas¬ 
ing the resistance the voltage is steadily increased until it gets to 
its maximum when the external circuit is open and the resistance 
is infinite, the whole voltage being then available for magnetizing 
the shunt field coils to their maximum strength. 

Fig. 5 shows the characteristic curves of a shunt-wound Gramme 
dynamo capable of giving 400 amperes at 120 volts. The arma¬ 
ture conductors were not capable of carrying more than 400 
amperes, and the part of the curves not actually found by experi¬ 
ment is shown in dotted lines. The lower curve marked E is the 
external characteristic while the upper curve marked E' is formed 
from it by adding to the external voltage at any point the corre¬ 
sponding value of the voltage lost in the armature at the given time. 




DIRECT CURRENT DYNAMOS. 


19 



Fig. 5. Characteristic Curves of Shunt-Wound Gramme Dynamo. 


























































































































































































































































































































































































20 


DIRECT CURRENT DYNAMOS. 


In order to have the magnetic circuit of a dynamo properly 
proportioned and dimensioned for the magnetic flux which it is to 
carry it is necessary, first, to find the necessary number of lines 
required to be cut by the armature conductors in order to produce 
the desired voltage, and then to provide a magnetic circuit of suffi¬ 
cient cross section for these magnetic lines. 

After a machine has been designed and built a magnetic sat¬ 
uration curve may be taken to test the condition of the magnetic 
circuit as regards saturation. This tests the accuracy of the cal¬ 
culations, and serves as a check. From an economical standpoint 
it is best not to carry the magnetic flux too near the point of sat¬ 
uration during normal working, for the number of ampere-turns 
required to produce a given magnetic flux is greater in a saturated 
field than in one not saturated. In certain types of machines, 
Lowever, there are certain advantages in practical points of working 
and regulation to be derived from having saturated fields that are of 
such importance as to outweigh the value of greater efficiency. 
By knowing from the type and the design of a machine to what 
degree its fields should be saturated, and by having an experi¬ 
mental saturation curve of the machine, a comparison can be 
male between the ideal and the result. Fig. 3 gives the satura- 
t on curve of the magnetic circuit of a Crocker-Wheeler dynamo 
for no load and with load, the dynamo running at a constant 
speed of 1,620 revolutions per minute. 

For a given field magnetizing force in ampere-turns the 
dynamo will show considerable difference in voltage between a 
run with load and one without. This shows quite plainly the 
effect of the lost voltage in the armature and also the loss due to 
the armature demagnetizing tendency. The values of voltage at 
the terminal of the machine were observed both for increasing and 
decreasing values of magnetizing current or ampere-turns. The 
result is shown in the double curves (see Fig. 3) and is caused by 
the phenomenon of hysteresis. The magnetism of the dynamo lags 
behind the increasing magnetizing force and so appears too low, 
and again it lags behind the decreasing magnetizing force and 
appears too high. 

The Internal and External Shunt Characteristic Curves of a 
Crocker-Wheeler dynamo are shown in Fig. 6. This machine is 



DIRECT CURRENT DYNAMOS. 


21 



0 
















































































































































22 


DIRECT CURRENT DYNAMOS. 


really a compound dynamo, and in order to get shunt character¬ 
istic curves, the series coils were simply disconnected and the ter¬ 
minals of the outside circuit were connected directly with the 
brushes. 

It will be noted from the figure that the voltage drops gradu¬ 
ally at first as the load increases and that it drops much more 
rapidly as the load reaches a maximum. By comparing this figure 
with Fig. 11 which shows characteristic curves fqr the same 
machine run as a compound dynamo, it will be noted that the max¬ 
imum load of the shunt machine is quite a little less than the 
maximum load of the same machine when the series coils are in 
use. The magnetizing effect of the series coils increases in 
strength practically in proportion as the load increases and so the 
series coils play a very important part in maintaining the strength 
of the field coils. It will be noted that the maximum energy as 
shown in the shunt characteristic curve is not quite 3J horse-power, 
while the maximum energy shown in the compound characteristic 
curve of Fig. 11 is nearly 8 horse-power. 

The speed at which the shunt characteristic curve was taken 
was 1,500 revolutions. The total current of the armature is 

i = /.+ j sh = i + A. 

^sh 

The total voltage generated by the armature is, 

E' = E+ j 1+ JL |r a . 

( r sh ) 

In the series-wound dynamo, there is but one circuit (see 
Fig. 7), and therefore but one current, 7, the value of which 



Fig. 7. Diagram of Series-Wound Dynamo. 

depends upon the value of the total voltage E' and the total resist¬ 
ance which is made up of r a , r se and R. The series field is com- 











DIRECT CURRENT DYNAMOS. 


23 


posed of a comparatively few turns of heavy copper conductor and 
is connected in series with the armature and the external circuit. 


The current may be expressed as 

r = /«„ = / 


*1 = 

-^se = 


r = 


E' 

R + r a + r se 


E + (E ! — E) _ E 

~R + ( r a + ^se) R 


r 


The total voltage may be expressed as 

E’ = E + 7(r. + r ae ) = E j 1 + + r “ 

f The electrical efficiency is 

_ useful energy _ EI _ E 
total energy E’l f E’ 

or in terms of resistances it may be expressed as 

_ i 2 R _ R 

Ve F 2 (B + r a + r se ) B + r a + r 8e 

Et 

From the equation/' = ——,- it is seen that an 

R + r a + r se 

increase in the outside resistance diminishes the current in the 
field coils, thus diminishing the magnetic flux. As the constancy 
of the magnetic flux depends upon the constancy of the current 
value these series-wound dynamos are best adapted to give a 
constant current, and are used mostly for running arc lamps on 
series circuits. 

From the equation E’ — E + I (r a -f- r se ) it is seen that 
the total current in the armature is the same as the current 
in the outside circuit, also that the total voltage is the external 
voltage plus the lost voltage in the armature and series coils due 
to the passage of the current. 

From the equation for electrical efficiency it is evident that it 
will be a maximum when the armature resistance and series coil 
resistance are as small as possible. 













24 


DIRECT CURRENT DYNAMOS. 



Fig. 8. 



































































































































































DIRECT CURRENT DYNAMOS. 


25 


Internal and external characteristic curves of a small Wood 
arc light dynamo are shown in Fig. 8. The total voltage generated 
by the armature is equal to the voltage observed at the terminals 
plus the lost voltage of the armature and series coils. 

This was calculated by the equation 

= E + I (r a + r ae ) 

where r a was 5.4 ohms and r se was 6.83 ohms. 

The observed voltages are corrected for a speed of 1150 
revolutions. 

The electrical efficiency for this machine was calculated and 
found as follows: 

DATA FOR EFFICIENCY OF SMALL WOOD DYNAMO. 


Speed 

I 

E f r 

El 

El 

E' r 

1150 

1 

288 

88 

31% 

1150 

4.4 

1727 

1408 

81% 

1150 

4.6 

1840 

1509 

82% 

1150 

6.0 

2704 

2280 

84% 

1150 

7.0 

8389 

2884 

85% 


The usual use of the constant current dynamo is for arc 
lighting. Arc lamps are commonly run in series and require 
about 10 amperes, and a potential of about 50 volts across the 
terminals of each. To maintain a constant current in the circuit 
the E. M. F. at the brushes of the generator must then increase 
50 volts as each lamp is thrown in. 

Constant current dynamos are series wound. Of such 
machines, we know that with a variable resistance in the external 
circuit, the current, or E. M. F. or both will vary. In the 
case of a machine supplying a certain current, if the resistance is 
increased the current strength will fall, and the field coils being 
in series the magnetic field is weakened, thereby lowering the 
E. M. F. and still further lessening the current. On the other 
hand if the external resistance is reduced, a larger current flows 
and a correspondingly higher potential is generated. Obviously, 
if such a machine is required to give a constant current through a 














26 


DIRECT CURRENT DYNAMOS. 


variable resistance, there must be some means provided for raising 
and lowering the E. M. F. 

For effecting such regulation there are these three methods: 
(1) Varying the speed. (2) Varying the field strength. (3) 
Changing the position of the brushes. 

The first method is seldom made use of although in principle 
it is very simple. There must be an automatic regulation of 
speed and the E. M. F. of course rises and falls with it. Such 
regulation, on account of the inertia of the heavy moving parts of 
the engine and generator, is too slow for lighting service. 

Regulation by the second method is obtained either by chang¬ 
ing the number of active conductors in the field or by changing 
the connections of the coils from series to parallel and vice versa. 
Running on light load (few lamps in series) only enough turns in 
the field are left in circuit to maintain a current of 10 amperes, 
and as more lamps are added to the circuit more field turns are 
thrown in to strengthen the field and to raise the voltage. The 
disadvantage of this method is the small range of loads that can 
be economically carried. For as the field coils are thrown in the 
cores soon rapidly approach the point of saturation, beyond which 
a large increase of ampere-turns effects the E. M. F. very slightly. 
On the other hand the machine will not run well on very light 
loads, for the current being constant the reaction of the armature 
is constant and on the weakened field has a greatly increased 
effect of distortion upon the field and causes serious sparking at 
the brushes. 

The third method that of shifting the brushes is the most com¬ 
mon one. To understand the principle of this regulation it will be 
well to refer back to the figure showing the two paths of the current 
through a ring armature and out by the brushes, (see Fig. 25 in 
Instruction Paper on Theory of Dynamo-Electric Machinery'). It 
will be seen readily that the maximum E. M. F. at the brushes is 
obtainable when they are at the neutral point, that is when all 
the amature coils on either side of the brushes generate an E. M. F. 
in the same direction. If the brushes are shifted from the neutral 
point then some of the coils on either side oppose the others on 
the same side and a reduced pressure at the brushes is the result. 
If the brushes were placed at points midway between the neutral 



DIRECT CURRENT DYNAMOS. 


27 


points, then on each side half the coils would oppose the other 
half and the pressure at the brushes would be nil, that is the 
algebraic sum of the E. M. F’s of the coils in each half would be 
zero. Then by moving the brushes from this point toward the 
neutral point this sum, or the E. M. F. of the machine would in¬ 
crease gradually to the maximum. Such is the principle of this 
method of regulation. 

A constant current machine running on light load will have 
its brushes in a position considerably oh the neutral point, and as 
more lights are thrown into the circuit making the load heavier, 
they will he brought correspondingly nearer, maintaining suffi¬ 
cient E. M. F. at the brushes to force a current of constant value 


E 

i 

R 



Fig. 9. Ordinary, or Short Shunt Compound Dynamo. 


through the increasing external resistance. The shifting of the 
brushes is done automatically by electromagnets in circuit, which 



Fig. 10. Long Shunt Compound Dynamo. 


E 

I 

R 


act instantly to counteract any change of current strength flowing 
through them. 

The Compound=Wound Dynamo (see diagrams Figs. 9 and 
10), may he considered as a shunt dynamo to which some series 
windings have been added to compensate for the fall in voltage at 


















28 


DIRECT CURRENT DYNAMOS. 


the brushes due to the demagnetizing effects of the armature, lost 
volts in armature, and to compensate for line losses as the load 
increases. The shunt circuit may either be connected to the 
brushes, in which case the machine is called the Ordinary or Short 
Shunt Compound Dynamo (see Fig. 9), or the shunt circuit may 
be connected to the terminals of the outside circuit in which case 
the machine is called a Long Shunt Compound Dynamo (see Fig. 
10). Using the symbols as marked on these diagrams the equa¬ 
tions for the ordinary or short shunt compound dynamo will be 
deduced. The total current is 

I' = I + J 8h = r S e + i* = i+ l + Ir « 

^*sh 


It may also be written (see similar equation on page 11) 


(ft + >V.e) X h 

-ft + r se + r sh 


or again it may be written 

1 _j_ R Hr r se 

( 7-sh 


_ 1 x r sh 4~ r se ~f~ R 
^sh 


The value of the shunt current may be written 

I sh = ^ 1 ^ _ j x ^se + R 

r sh r sh. ^ S h 

t The value of the total E. M. F. generated may be written 

E'.~E+I\r* + Ir M 

* = JE + D+ —+ Jr ” k+JV se 

^ ^*sh ' 

f The value of the total E. M. F. generated may also be 
expressed in terms of the useful voltage and the various resistances. 
It is equal to the sum of four lost voltages. The first of these 
voltages is E, the useful voltage. The next is the voltage lost in 

the series coil and is equal to X E. 

R 

Then the total voltage at the brushes is 

E + E or E ] 1 + | or e\ R + I 

-ft ( R ) I R < 













DIRECT CURRENT DYNAMOS. 


29 


The lost voltage in the armature clue to that part of the cur¬ 
rent that flows through the series and outside circuit is to the 
voltage of the series and outside circuit as the resistance of the 
armature is to the resistance of the series and outside circuit. 

Lost volts in armature # ^ j R -f- r se 
due to outside current * j R 

Lost volts in armature _ ^ J r a (R -\- r se ) £ 

due to outside current — f Rlx R ) ( 

Similarly the lost voltage in the armature due to the current 
flowing in shunt field is to the voltage of the shunt field as the 
resistance of the armature is to the resistance of the shunt field. 
Lost volts in armature # j R H~ r se 

due to shunt current * ) R 



- r a 


~b R 


Lost volts in armature _ E (R + r se ) r a 
due to shunt current — R X r sh 

Therefore, the total voltage generated being equal to the sum 
of these four expressions it may be written 

Jjir R^se E (R + ^se) ^*a l C R “H R 

~R~ R (r Be + R) r ah X R 

As the third term when simplified has the same denominator 
as the second term the two may be combined, and as all four terms 
in the right hand side of the equation contain E it may be taken 
outside the parenthesis j then, 

E' = E I 1 ^ se ~ l~ r & I r a (R ~~h r se) ) 

\ R R X r sh \ 

The electrical efficiency of the ordinary compound dynamo is 
equal to the useful energy divided by the sum of the useful energy 
in the outside circuit, plus energy lost in series coils, plus energy 
lost in shunt coils, plus energy lost in armature. 

El _ PR 

Ve ~ WT - P II + R r se + / 8h 2 + I' 2 r a 


P R 


R (jse ~l~ r sh ~l~ Ry^r a _j_ IJi ^ t -)- P R -f- Pr t 

f S h 2 r s h 


t = 


Ose + ^sh + _|_ O'se + R ) 2 I r se 

r^-R T r sh 7? ^ ^ R 






















80 


DIRECT CURRENT DYNAMOS. 


For the Long Shunt Compound=Wound Dynamo (see Fig. 
10) the equations may be written in a similar manner. The 
values of the total current generated will be 


I' = I se = I+I sh =I + 


E 

r S h 


= I 




sh 


Ish 


E_ _ 

^*sh ^sh 


The total voltage generated in the armature is 


E' = E+I' (r a + r se ) = E+ j 7+ A l (r . + r se ) 

t ^Sh ) 


• = x]l + X+?*\ (r a +r 3e ) 

( E r A ) 

f*The electrical efficiency of the long shunt compound dynamo is 
_ I- R _ 

I' 2 ( r a + r se) + -4h 2 r sh~\- I 2 R 

P R 

TRI* + R* js 

'•sh '•sh 2 


I ('•a + '-se) + I*R 

) r.v. 



t = 


' - a + ' , ! 

R 


+ 2 


+ r t 


sh 


R ('•a 


^*sh" 


r se) | R | q 
^sh 


The compound-wound dynamo by means of its shunt coils and 
its series coils is enabled to produce a practically constant voltage 
within the working limits of its capacity. This is not absolutely 
true for the voltage curve is a slightly curved line tending to rise 
slightly from no load to about half load and then falling again 
from about half load to full load. 

As the load increases and the resistance of the outside circuit 
decreases the main current in the series field coils increases and 
the current in the shunt coils still remains constant as it is now 
fed with a constant voltage. Thus although there are more lost 
volts in the armature to cut down the voltage, and although there 
is a much greater demagnetizing effect produced by the armature 
yet the voltage supplied to the external circuit is still maintained 
at its constant value. 














DIRECT CURRENT DYNAMOS. 


31 


As the load decreases and the resistance of the outside cir¬ 
cuit increases less current flows through the series coils and 
through the armature, and as there are less lost volts in the armature 
and a less demagnetizing effect it is readily seen that the magnet¬ 
izing effect of the series coils is needed less now, and the shunt 
field coils which are still supplied at about the same constant volt¬ 
age play relatively a much greater part in furnishing the magnetism 
of the dynamo. A compound dynamo therefore if properly pro¬ 
portioned, will supply a practically constant voltage at all loads. 
In the case of the ordinary or short shunt compound dynamo, the 
potential at the brushes is kept constant. In the case of the long 
shunt dynamo the potential at the terminal of the working circuit is 
constant. The latter arrangement therefore is somewhat preferable 
but either arrangement proves satisfactory in well designed dynamos 
as the actual value of the difference in the two cases is generally 
very slight. 

In the case of the ordinary or short shunt compound dynamo 
the series coils furnish the excitation required to produce a potential 
such as will compensate for the lost voltage in the armature and 
the demagnetizing effects due to the armature. In the case of the 
long shunt dynamo the series coils compensate for these same 
losses as well as for the lost voltage of the series coils. 

It is advisable and generally customary to put a somewdiat 
greater number of series turns on the field coils than is necessary 
to overcome armature reactions and lost voltage in order to have 
the dynamo give a somewhat greater voltage at full load than at 
no load. This process which is called overcompounding is calcu¬ 
lated for a rise of voltage of about four or five per cent. As the 
load increases an engine often runs a few revolutions slower, or 
there is a trifle more slipping of the belt which causes the speed 
of the armature to drop a little; the increasing load is also accom¬ 
panied by a greater lost voltage in the line or feeders ; hence the 
dynamo should be overcompounded to make up for these losses. 
If the load is at some distance from the generator there might be 
considerably more than 5 % drop of voltage at full load. This drop 
can be figured and a machine may be especially compounded for 
this loss at the time it is built. The object of course is to keep 
the voltage constant at the lamps, and if they are some distance 




32 


DIRECT CURRENT DYNAMOS. 


away and the load is constantly changing it is often necessary and 
advisable to run so-called pressure wires from the center of dis¬ 
tribution back to a voltmeter in the dynamo room. Then if the 
dynamo is not overcompounded so as to give the proper pressure at 
all loads the pressure may be varied by adjusting the rheostat 
which is in series with the shunt coils of the dynamo. Even 
though the dynamo is properly compounded it is almost always 
necessary to regulate with the rheostat also in the case of incan¬ 
descent lighting, especially where the lamps are distributed over a 
wide territory. 

In designing a dynamo for any given output it is necessary to 
make due allowance for the total energy to be generated which is 
the total voltage U 1 multiplied by the total current I'. In order 
that the energy lost in the series field turns may not be too great, 
it is well to have them wound near the field cores so that each turn 
may be as short as possible and then the shunt coils may be wound 
outside of the series turns. 

In Fig. 11 are shown some characteristic curves of a compound- 
wound Crocker-Wheeler dynamo running at 1500 revolutions per 
minute. 

The upper curve is the external characteristic of the dynamo 
running with all the resistance cut out of the shunt field regulat¬ 
ing rheostat. It is running, therefore with a maximum field exci¬ 
tation and giving its maximum voltage, about 127.8 volts at no 
load. It will be noted that as the load increases the voltage drops 
somewhat. It is evident that the magnetic flux due to the series 
ampere-turns on the field coil is not great enough to make up for 
the armature demagnetizing effects and the lost volts in the 
armature. Therefore there are not enough series turns, and the 
dynamo is undercompounded when running at 127.8 volts. 

Next some of the resistance of the regulating rheostat is put 
in series with the shunt field until the voltage falls to 109.2 volts. 
By increasing the load the voltage increases, being from about 4 % to 
7 %‘higher as the load increases than it was at no load. Now the 
rise in voltage, due to compounding, makes up for the loss in the 
line wires and is a little too great in the case of the 7 % ,and in order 
to keep the lamps from burning too brightly the rheostat handle 
would have to be turned so as to cut the voltage down a little. 





DIRECT CURRENT DYNAMOS. 


33 



Fig. 11. Characteristic Curves and Efficiency Curve of a Crocker-Wheeler 
Compound Dynamo. 














































































































34 


DIRECT CURRENT DYNAMOS. 


Judging fro in these two curves it appears that if the dynamo were 
run at a little higher initial voltage, say 112 or 115 it would be then 
running at just about the jDroper voltage for which it was com¬ 
pounded. The percentage of the overcompounding would then be 
about four or five per cent and the dynamo might run at varying 
loads with little or no need of adjusting the rheostat. 

The curve shown by broken line (see Fig. 11) is the internal 
characteristic curve derived by adding the values of the lost volts 
in armature and in series field to the value of the external voltage. 

The lower compound characteristic curve starting with an ini¬ 
tial voltage of 89 shows that the dynamo is very much overcom¬ 
pounded when running at this voltage. The rheostat has been so 
adjusted that less current flows through the field coils, the shunt 
current is reduced and the value of the shunt ampere-turns is 
reduced. On the other hand the value of the series ampere-turns 
is as great as formerly as the load increases, and so the value of- 
the series excitation is now large in comparison with that of the 
shunt. The voltage therefore increases greatly with the load, 
causing the dynamo to be greatly overcompounded when started 
at an initial voltage of 89. 

In Fig. 11, is also given a curve of commercial efficiency for 
this 5 H. P. Crocker-Wheeler compound dynamo, the commercial 
efficiency being the useful electrical work, E I, derived from the 
dynamo, divided by the value of the mechanical work delivered to 
the pulley of the dynamo. 

The following table contains the data from which the charac¬ 
teristic curves shown in Fig. 11 were drawn. 


I 

E 

1 

I 

E 

I 

E 

0 

127.8 

0 

109.2 

0 

89. 

3.6 

128.0 

3.4 

110.0 

2.7 

89.7 

8.4 

128.0 

8.7 

112.0 

7.2 

91.3 

14.2 

127.7 

13.2 

115.0 

11.3 

96.4 

18.4 

127.7 

17.1 

117.0 

15.6 

101.8 

23.8 

127.6 

22.6 

117.0 

20.6 

105.0 

27.2 

127.2 

25.6 

116.8 

24.6 

105.4 

30.9 

126.0 

29.3 

116.0 

27.9 

107.5 

35.2 

124.0 

33.7 

116.0 

32.0 

108.2 

39.3 

122.9 

37.1 

115.9 

36.1 

108.0 

43.8 

121.9 

42 8 

114.0 

40.3 

107.0 





40.4 

107.0 


The characteristic curves of a 5 KW„ Edison compound 


















DIRECT CURRENT DYNAMOS. 


35 



Pig. 12 . Internal and External Characteristic Curves and Efficiency Curve of 
5 KVV. Compound Edison Dynamo. 



































































36 


DIRECT CURRENT DYNAMOS. 


dynamo are given in Fig. 12, for initial voltages of 131, 110, and 
90, at a speed of 1,700 revolutions. In testing dynamos it is very 
difficult to keep the speed absolutely constant. There may be a 
variation in the speed of the main pulley from which the dynamo 
is run, and the belt may slip more and more as the load increases. 
However, as the voltage is proportional to the speed, the observed 
voltage may be corrected for the observed speed quite readily. 
The reading of the voltmeter and the tachometer should both be 
taken at the same instant. 

The characteristics of the compound Edison dynamo as 
shown in Fig. 12, present very much the same phenomena as 
were exhibited by the characteristics of the Crocker-Wheeler 
dynamo in Fig. 11. The Edison dynamo, the curves of which are 
shown, is probably properly overcompounded for about 4 % or 5 % 
when running at an initial voltage of about 115. A curve of elec¬ 
trical efficiency is also shown. 

The following table gives the data for the characteristic and 
efficiency curves of the 5 KW. compound Edison dynamo. 


Speed. 

E 

E corrected 
to 1,700 rev. 

I 

Speed. 

E 

Corrected 

E 

I 

E I 

E'l' 

1740 

131.5 

128 

0 

1720 

110 

108.1 

0 

0 

1725 

130.0 

128.1 

3.8 

1715 

111 

109.3 

8.7 

.82 

1730 

130.0 

128.3 

9.8 

1690 

111 

111.0 

22.0 

.86 

1725 

128.5 

126.7 

14.7 

1670 

110 

111.3 

28.7 

.87 

1720 

128.0 

126.6 

19.2 

1665 

112 

113.1 

37.3 

.82 

1715 

126.5 

125.4 

24.3 

1660 

112 

114.0 

47.5 

.80 

1710 

126.5 

125.6 

28.0 

1730 

91.5 

90.0 

.0 


1705 

125.5 

125.3 

32.2 

1730 

91.5 

90.0 

7.4 


1685 

124.0 

125.2 

36.0 

1725 

93.0 

91.6 

14.6 


1670 

123.0 

125.3 

40.3 

1715 

96.5 

95.7 

22.4 


1665 

123.0 

125.6 

45.0 

1685 

99.0 

99.9 

30.0 


1635 

118.5 

123.3 

46.5 

1660 

99.5 

101.9 

38.0 






1640 

100.0 

103.7 

45.4 



















DIRECT CURRENT DYNAMOS. 


37 


CALCULATION OF DIRECT CURRENT 
DYNAHOS. 


Note.— In the following pages will be found the electrical and 
magnetic details for the design of a 50 KW. shunt-wound dynamo. 
In designing each part, first the physical phenomena are discussed ; then 
the equation is given ; and then the calculation is made. These treat¬ 
ments of the physical phenomena should be carefully studied, especially 
those treating on the magnetic circuit appearing on pages 56 to 58. 
It is not so important to make the actual calculations as it is to under¬ 
stand quite thoroughly the phenomena underlying the working of the 
machine. However, it should be quite helpful to trace through the 
calculations if one is interested. The mechanical design of the dynamo 
will not be treated. The table of symbols is not to be studied but is 
simply arranged for easy reference. It is based on standards of nota¬ 
tions, and is recommended for use by any who may become interested 
in dynamo design. 


LIST OF SYHBOLS. 

AT, at — ampere-turns. 

AT = total number of ampere-turns on magnets, at normal load, 
or magnetizing force. 

AT' = total magnetizing force required for maximum output of 
machine. 

AT" = total magnetizing force required for minimum output of 
machine. 

at a = magnetizing force required for armature core, normal load. 

at cim = magnetizing force required for cast iron portion of mag¬ 
netic circuit, normal load. 

at cmBm = magnetizing force required for cast steel portion of mag¬ 
netic circuit, normal load. 

at g = magnetizing force required for air gaps, normal load. 

at m = magnetizing force required for magnet frame, normal 
output. 

at v , at Vo = magnetizing forces required for polepieces. 

at r = magnetizing force required for compensation of armature 
reactions. 

at wL = magnetizing force required for wrought iron portion of 
magnetic circuit, normal load. 






38 


DIRECT CURRENT DYNAMOS. 


at y , at Yo = magnetizing forces required for yoke or yokes. 
a = half pole-space angle (also angle of brush displacement). 

(ft = magnetic flux density in magnetic material, in lines per 
square centimetre. 

(ft" = magnetic flux density in magnetic material, in lines per 
square inch. 

(ft a > (B" a = average density of magnetic lines in armature core. 
(ft ai . (&" ai = maximum density of magnetic lines in armature core. 
(& a , (ft " a2 = minimum density of magnetic lines in armature core. 
(ft ci , (B" c .i. = mean density of magnetic lines in cast iron portion 
of frame. 

&cs.’ c.s. = mean density of magnetic lines in cast steel portion 
of frame. 

& p , (ft" p == mean density of magnetic lines in polepieces. 

(ftp , (ft" Pi = maximum density of magnetic lines in polepieces. 

(ftp , (ft " p2 = minimum magnetic density in polepieces. 

(ft w i, = magnetic density in wrought iron portion of mag¬ 

netic circuit. 
b — breadth, width. 

6 a = breadth of armature cross-section, or radial depth of arma¬ 
ture core. 

b\ = maximum depth of armature core. 

b y = breadth of yoke. 

ft = angle embraced by each pole. 

/3 r x = percentage of effective arc, or effective field circumference. 
7 = electrical conductivity, in mhos. 

_Z), d, 3 = diameter. 

D m = external diameter of magnet coil. 

d a = diameter of armature core. 

d ! a = mean diameter of armature winding. 

d'\ = external diameter of armature (over winding). 

d"\ — mean diameter of armature core. 

d c = diameter of core-portion of armature shaft. 

d f — mean diameter of magnetic field. 

d m = diameter of magnet core. 

d v = diameter of bore of polepieces. 

3 a = diameter of armature wire, in mils. 

8 " a = height of insulated armature conductor, in inches. 



DIRECT CURRENT DYNAMOS. 


39 


= thickness of iron laminae in armature core, in inches. 

& m diameter of magnet wire, hare, in mils. 

S' m = diameter of magnet wire, insulated, in mils. 

8 se == diameter of series field wire. 

3 sh = diameter of shunt field wire. 

(3 a ) 2 = sectional area of armature conductor, in circular mils. 

E, e — electromotive force, or pressure, in volts. 

E = normal E. M. F. output, or voltage of generator; terminal 
E. M. F., or supply voltage of motor. 

E' = total E. M. F. induced in armature of generator; counter 
E. M. F. of motor. 

e = unit armature induction per pair of poles, volts per foot. 
e' = specific induction of active armature conductor, volts per 
foot. 

e a drop of voltage due to armature resistance. 

e = factor of eddy current loss in armature, English measure 
(watts per cubic foot), 
e = eddy current constant. 

— magneto-motive force, in gilberts. 

F, f = force, or pull, in pounds. 

/ ((B) = function of (B; magnetizing force per centimetre length 
for density (B. 

f (&") = function of (B" ; magnetizing force per inch length for 
density (B". 

/ (®a)?/ (® /r a) — specific magnetizing force of armature core. 
/(®ci)?/ — specific magnetizing force of cast iron por¬ 

tion of magnetic circuit. 

/ (® P )i/ (®%) — specific magnetizing force of polepieces. 

/ (®w.O’ = s P ec ifi c magnetizing force of wrought iron 

portion of magnetizing circuit. 

/ (® y ),/ (CB" y ) = specific magnetizing force of yoke. 

O = useful flux, i. e., number of lines of force cutting armature 
conductors, at normal output. 

<£/ __ total flux, or total number of lines generated, at normal 
output (webers). 

<|>" == total flux per magnetic circuit. 

5C = magnetic flux density in air, or field density, in gausses 
(lines of force per square centimetre). 



40 


DIRECT CURRENT DYNAMOS. 


3C" = field density, in lines of force per square inch. 
h = height, thickness. 

7 i a = total height of winding space in armature (depth of slots). 
A' a = available height of armature winding space. 

A m = height of winding space on field magnets. 
hJ m = net height of field winding. 

7i p = height of polepieces. 

A y = height of yoke. 

HP , hp — horse power. 

7 ] — factor of hysteresis loss in armature, English measure (watts 
per cubic foot). 
j] 1 — hysteretic resistance. 
rj c — commercial efficiency. 

7j e = electrical efficiency. 

/, i — intensity of current, amperes. 

I — current output, or amperage, of generator; current supplied 
to motor terminals. 

1' — total current active in armature. 

I m — current in magnet winding. 

I se = total series current, in amperes. 
i" sh total shunt current, in amperes. 

7 a i= current density in armature conductor, circular mils per 
ampere. 

i m = current density in magnet wire, circular mils per ampere. 
i se ~ current density in series wire, circular mils per ampere. 

7 sh — current density in shunt wire, circular mils per ampere. 

K, k = constants. 

Ai, A 2 , A 3 , . . . = various constants depending upon material, etc. 

L, l — length, distance. 

L a — active length of armature conductor. 

L e = effective length of armature conductor. 

L m — total length of magnet wire, in feet. 

Z sh = total length of shunt wire, in feet. 

L t — total length of armature conductor. 

7 a = length of armature core. 

l’\ — length of magnetic circuit in armature core. 

V c \. — length of magnetic circuit in cast iron portion of field 
frame. 




DIRECT CURRENT DYNAMOS. 


41 


l" g = length of magnetic circuit in air gaps. 

Z p — length of polepieces, parallel to armature inductors. 

Z' p = mean distance between pole-corners. 

Z" p — length of magnetic circuit in polepieces. 

Z t — mean length of turn of field magnet winding, in feet. 

I" r =■ length of mean shunt turn. 

Z" w i# — length of magnetic circuit in wrought iron portion of field 
frame. 

Z' y = length of yoke. 

I" y — length of magnetic circuit in yoke. 

X = factor of magnetic leakage. 

X m = — = specific length of magnet wire, in feet per ohm. 

Pm 

M] M x , . . . — mass, volume. 

M = mass of iron in armature core, in cubic feet, 
yu, = magnetic permeability 
N, n — number. 

N — number of revolutions of armature per minute. 

N' — number of revolutions of armature per second. 

N x = frequency of magnetic reversals, or number of cycles per 
second. 

iV a r= total number of turns on armature. 

N c — number of conductors around pole-facing circumference of 
armature. 

N m = number of turns on magnets. 

N se = number of series turns. 
iV sh = number of shunt turns. 

— number of turns per armature coil. 
n c — number of armature coils, or number of commutator divisions. 
n x = number of layers of wire on armature. 
n v — number of pairs of magnet poles. 

n' v — number of pairs of parallel branches in armature, or number 
of bifurcations of current in armature. 
n w = number of armature wires per layer. 
n z = number of magnetic circuits in dynamo. 

% . . . = permeances. 

^ = relative permeance of gap-spaces. 

= relative average permeance across magnet cores. 



42 


DIRECT CURRENT DYNAMOS. 


— relative permeance across polepieces. 

= relative permeance between polepieces and yoke. 

= relative permeance of clearance space between poles and 
external surface of armature. 

P' = total electrical energy, active in armature, or electrical 
activity of machine. 

P" — mechanical energy at dynamo shaft; i. e ., driving power of 
generator, output of motor. 

P A ~ total energy absorbed in armature. 

P M = total energy absorbed in field circuits. 

P a = energy absorbed in armature winding (J 2 P loss). 

P e ~ energy absorbed by eddy currents, in watts. 

P h — energy absorbed by hysteresis, in entire armature core. 

P m = energy absorbed in magnet windings. 

P Q — energy loss due to air-resistance, brush friction, journal 
friction, etc. 

P' Q == energy required to run dynamo at normal speed on open 
circuit. 

P se = energy absorbed in series winding. 

P sh — energy absorbed in shunt winding. 

7r = ratio of circumference to diameter of circle, = 3.1416 or - 2 y 2 - 
nearly. 

(R = reluctance of magnetic circuit, in oersteds. 

P, r = electrical resistance, in ohms. 

P = resistance of external circuit. 

R a = total resistance of armature wire, all in series. 

r a = armature resistance, cold, at 15.5° Centigrade. 

r\ = armature resistance, hot, at (15.5 -|- ^ a ) degrees Centigrade. 

r m = magnet-resistance, cold, at 15.5° Centigrade. 

r' m = magnet-resistance, warm, at (15.5 -j- # m ) degrees Cent. 

r sh — resistance of shunt winding, cold, at 15.5° Centigrade. 

r' sh = resistance of shunt winding, at (15.5 —J- # m ) degrees C. 

p m = resistivity of magnet-wire, in ohms per foot. 

S = surface, sectional area. 

S A = radiating surface of armature. 

S & = sectional area (corresponding to average specific magnetizing 
force) of magnetic circuit in armature core. 
jS &1 = minimum cross-section of armature core. 



DIRECT CURRENT DYNAMOS. 


43 


$ a2 = maximum cross-section of armature core. 

S c i — sectional area of magnetic circuit in cast iron portion of 
field frame. 

S C S ' — sectional area of magnetic circuit in cast steel portion of 
field frame. 

S { — actual field area; i. e., area occupied by effective inductors. 
S g — sectional area of magnetic circuit in air gaps. 

— radiating surface of magnets. 

S m = sectional area of magnet-frame, consisting of but one 
material. 

N p = area of magnet circuit in polepieces of uniform cross-section, 
tfpi = minimum cross-section of polepieces. 

Np 2 == maximum area of magnetic circuit in polepieces. 

— sectional area of magnetic circuit in wrought iron portion 
of field frame. 

S y = area of magnetic circuit in yoke. 
a — factor of magnetic saturation. 

T,t — time. 

r = torque, or torsional moment. 

0 a rise of temperature in armature, in degrees Centigrade. 

0 m = rise of temperature in magnets, in degrees Centigrade. 
v = velocity, linear speed. 

v c = conductor velocity, or cutting speed, in feet per second. 

W t , wt = weight. 

wt\ — weight of armature winding, covered wire. 
wt / m = weight of magnet winding, covered wire. 
wt' se = weight of series winding, covered wire. 
wt' sh = weight of shunt winding, covered wire. 



WIRE TABLE. PROPERTIES OF COPPER WIRE. 


44 


DIRECT CURRENT DYNAMOS 


Temperature 

Coefficients. 

•suiqo 
! IOOjITIJM 
jad 

souRisisoy; 

22 

1.00000 

1.00387 

1.00776 

1.01166 

1.01558 

1.01950 

1.02343 

1.02738 

1.03134 

1.03531 

1.03929 

1.01328 

1.04728 

1.05129 

1.05532 

1.05935 

1.06339 

1.06745 

1.07152 

1.07559 

1.07968 

1.08378 

1.08788 

1.09200 

1.09612 

1.10026 

1.10440 

1.10856 

1.11272 

1.11689 

1.12107 

1.16332 

1.20625 

1.24965 

1.29327 

1.33681 

1.37995 

1.42231 

•3 -S3Q 
m ‘auiax 

OHNCO'tlOWt'OOa)0"NM^ , L'J«t.«fflO-'N«'tiOOI> l »fflOOQOOOOO .. 

H T-triHT-iriririi-ir-ir-iC^CqC^cqClCMiM <M.<N (N CO r}< ^ CO CO C5 O 

0* —* . 

•paaaAOQ 1101103 
•uj ‘uiq .lad stunjL 

t6‘occoqe|-rC)iftOwt)oioot^Nqo«r;r.ioNCO'j'Nwooo^"-'HOf-«r.t»wooi'0 

O T Hi-ir2c4cococoTr»cia«ot»t-ooo5r-'e4coio«ct2as(M't2r2^Joo< m'j> cit2t8oc2rj<c4r-oio500'—0t2 

Safe Cur. 
jAmperes 

•Jly 931 jJ Ul 

3p*ta 

dOQ'SOddC.OOOOOCCC 00 OJ 1 — !N rt C- <M 

® to c4 t2 05 00 t2 00 r-.' od to <0 00 ' r-T 10 Q lS <35 to Tt>* © 05 

»Hr+i©f©C-'J<<M©©t-©lOTTfCOcOOI<N» : -.T-.?.^;5 . 

CO C4 5) rH ri r-p rH 

•papuEfj 

ay iqsua 

©o©®oo©o©©©oci05 05co©©cooqTj;oi(Moj 

®tot>©ioc8c4coiSooc4t2c4ooTi<r-io5t2iacoi-<©05odi>. 

•HTf'Mr-<© 00 t-©lO'*t'Tt<e 0 C 0 C'S<M<M,--<'-CT-ll-i,-ir^ . . . . . ... . . . . . 

hi M • 

c c 0 

O *** O 

-pjBR 

t' , 3§§3©S33inS3iSi.';3irt<Ne3c3c3c'je3<Sc3S3SSS3. 

^ •' • • • • • • • • • • • • • • 

# p9[B3U 

-uv 

_iV5iSiSLOC'i©toco©ootDTfT-'05G-tot»<etjoj'-i©05o6t>tbiSiSiO>S 

^^^Tf^WCOCOCOCqC^lMCQrirHTHr^rHrH^rH . 

Tensile 
Sgth. in 
Pounds. 

“P iB H 

. 

rHC5©(M©©^-'Tt?©lOC'J©t>©'^ l COCO(NCIr^r-i . 

1 Ci l>» rf CO CO 03 t-h rn ih 

•p9lE3U 

-uy 

« Si$gg!3333SSS88iSS£8833g§388c;2S2 . 

»H O -JJ LO QO M i> 4 ^ GO lO CO Cl CM rn r-i m , . , . .. 4 . 

Resistance in International Ohms at 68 ° F. 

According to 

Matthiessen’s Standard of Resistivity. 

Feet per 
Ohm 
Ai- 
nealed. 

t—COiO’-’CiiCTfLOCi’-fiA 

coNrj^ic^coc5r}jco(^q(Nqo5 0 rt ^^t^o^oboqo^oitocSo} 

*3 ooQOOOrHfNLOHLOiOfOLOOQcO’taiNco^ocoociajddifdociait^QTtcocodiHf-lTH 
rH ^’HiSocoHcocoajWTHOi^QajcoQOrH^ciLOCsiot^^TPOcocqf-i^rt - 
VW05C5Q1'QQHinQlO^O^COi3cO«^nrtr( 
gO^OOOCOlS^COC^C^THT-lTH 

Ohms per Mile. 

Hard- 

Drawn. 

O N CO O C5 

. .*. 

«i’l! c ^Oa5r~C'l'M 05 00 05>0lQC0Tft0f / 5Tt-cqWt0-f<t0©05 

«'-6©t'iC'Jto-t‘©tooocroooit^oooo©©^.©-i;C5 —'©-r. 

H(N^-flOO°qC c OOriC3cO^^NCl';«O^CDj4'frfb. * . 

.. ^ r-5c4c4cd^idsoo6©eot^i-«t>'*c'S ! i? . . ’ .•. 

rHr-lHC^O^CO’tlO 

An¬ 

nealed. 

11 

.25835 

.32577 

.41079 

.51802 

.65314 

.82368 

1.0386 

1.3094 

1.6516 

2.0825 

2.6258 

3.3111 

4.1753 

5.2657 

6.6369 

8.3741 

10.555 

13.311 

16.785 

21.168 

26.691 

33.655 

42.441 

53.539 

67.479 

85.114 

107.29 
135.33 
170.59 
215.16 

271.29 
242.09 
431.37 
543.84 

685.87 

864.87 

1090.8 

1375.5 

1734.0 

2187.0 

2757.3 

3476.8 

4384.5 

5528.2 

Ohms per 1000 Ft. 

Hard- 

Drawn. 

O Tt< 00 

COQ^OClCO^HNW'f l-CO _ _ .. 

0 0 0 0 ^ 10 h 0 co co d t- ai ^ 00 m ai co 0 0 «o „ 

oocoosooajHCoajcoaDrHcoaiiOTHTfNOajCJoooici 

3 UJONONiOOWnQp^O»HOOM^M005»HrtN. , . 

^©©©r-jrM^HC^joito^Lotooqocj©©©^!©^!^)^;© 1 

‘ ^ rJ h d co ^ 10 6 co 6.. \.’. 

An¬ 

nealed. 

COOOh 

O^COHNOt-OCC^C^HCOM 

COHNCCCCOOCCCl^l-t^O, 

'fcot-aicnoci'fr-oiciC'iaiOji^cooci^o^Nco^ooc'U^ccir-inL^o^ 
a ooooHHriNwwjjj«5^0ieiiooiiOHqo i coOrtNcoM.«i5wNcoi'jNqc5oqo-iq^c<jo^o 

'i-^T-4T-^c4c0^iOSOa6oc4cDOlCC<io~*Tj!^eOcSc'5<OQOQT£C'$odQt'* 

Weight per Unit of. Length. 
Specific Gravity, 8.89. 

Ohms per 
Pound, 
Annealed. 

8 

.00007639 

.0001215 

.0001931 

.0003071 

.0004883 

.0007765 

.001235 

.001936 

.003122 

.004963 

.007892 

.01255 

.01995 

.03173 

.05045 

.08022 

.1276 

.2028 

.3225 

.5128 

.8153 

1.296 

2.061 

3.278 

5.212 

8.287 

13.18 

20.95 

33.32 

52.97 

84.23 

133.9 

213.0 

338.6 

538.4 

856.2 

1361 

2165 

3441 

5473 

8702 

13870 

22000 

34980 

Feet 

per 

Pound. 

rHClC'iONNCO’tO 

lO ^ Ci Ci Cl CS O LO CO O CN °0 r-H O t- 05 CO 00 ^ O 

HrH^cIcCTtiCOOOOMOOLO'MgH^rHMOCJOQOQrJ-WCi’-OpfHCfiHfiH 

ntH.ri^iMCO^iCCDOOOCOOOOClH^L^M^^eOiS^ 

r^rHr-C'KMCO^LOcioOOCOCD.rHgrt 

Pounds 

per 

Mile. 

rHHOb-COCLOrt'COOOC'lNr-if-rH^incO'tO'-'iO^Q^THCiCO 

ajciOLOi.oaj^icwacsooNcccSai^^erMocicociiftC'io^rHOrtoOHioaS 

^C100Q^C0OC0Oi0F : 00qC5«0C0^^O^^0iiqC0 05C|T-<^THqClOq«00|q00OiC’<f 

i-5 c-i 0 cd t> 0 h ci 0 <m* cd q> id ^ c4 cd rnVi \6 © © <n’ 0 06 0 16 cd c4 c4 1-1 i-i th.* * 

OOCOC<JCOCO r 0^t'CD(M^COCOQCl5cOOOOI>lO'«fCOC4C4r-ir^»-i 

COOHOCOOOO^lC'tW^eiHTHrt 

CO rH r-t »—• 

Pounds 
per 1000 
Feet. 

CiC<JOt^OOlCOCOI>-fO^$oSSS§ 

,OOOOOt^OiiM(NLO(NMC5CiOC5CO^fN»Hf-.OiOiOQt-hS 

0(NOOCOCOCOt^OO^LO^Q,i-iC5CiiOT^^tc4cOCDT-HOOCOO'»tCiOC^ait>-^Tt<fOo4 

10 icocciccooicori<ci^qc5co^a5i^o^oqooe4'a5ooo^aiiA(NOiL^^TfcooqrjrjiMqqqooo 

ocdoia5cdQc^oQaicoc5airH^a5idr^ait>o^cdcoc4r-iT-ii-i. 

-HQOHiflQOC^Ob-CO^COCOCqHHn 

O lO-* CO 04 05 ^ *- r- 

Area in 

Thou¬ 
sandths of 
an Inch 
d 2 x.7854. 

4 

.166190 

.131794 

.104518 

.082887 

.065732 

.052128 

.041339 

.032784 

.025999 

.020618 

.016351 

.012967 

.010283 

.0081548 

.0061656 

.0051287 

.0040672 

.0032254 

.0025579 

.0020285 

.0016087 

.0012757 

.0010117 

.00080231 

.00063626 

.00050457 

.00040015 

.00031733 

.00025166 

.00019958 

.00015827 

.00012551 

.000099536 

,000078936 

.000062599 

,000049643 

.000039368 

.000031221 

.000024759 

.000019635 

.000015574 

.000012345 

.0000097923 

.0000077634 

Circular 

Mils 

d 2 . 

3 

211600.00 

167806.43 

133076.66 

105534.50 

83692.67 

66371.31 

52634.37 

41741.32 

33102.37 

26251.37 
20818.35 
16509.64 
13092.75 
10383.02 

8234.11 

6529.95 

5178.48 

4106.72 

3256.78 

2582.74 

2048.29 

1624.30 
1288.13 
1022.53 

810.12 

642.45 

50£.49 

404.64 

320.42 

254.10 

201.52 

159.81 

126.74 

100.51 

79.70 

63.20 

50.13 

39.75 

31.52 

25.00 

19.83 

15.72 

12.46 

9.88 

•ipui 100* 

= HIM I S UIM 
u ! uP„ wiaureia 

2 

460.000 

409.642 

364.796 

324.861 

289.296 

257.626 

229.422 

204.307 

181.941 

162.022 

144.285 

128.490 

114.434 

101.897 

90.743 

80.808 

71.962 

64.084 

57.069 

50.821 

45.257 
40.303 
35.890 
31.961 
26.463 
25.346 
22.572 
20.101 
17.901 
15.940 
14.196 
12.642 

11.258 
10.025 

8.928 

7.950 

7.080 

6.305 

5.615 

5.000 

4.453 

3.965 

3.531 

3.145 

•aSnEQ 

•s % , a. J 3 C l uin N 

0 0 0 0 ih n co ^ iq co c- co 05 0 ih <?q n ^ 10 co t» oo © 0 rH <n n m co c- ao © o i-t co t(< uo co t«* qo ® o 

_ OOO rHrHrHrHrHr-irHrHrHfHC4C4C4C4G^C404C4C4C4COCOCOCOCOCOCOCOOQCOT^ 

H OO 

O 









































































DIRECT CURRENT DYNAMOS 


45 


WIRE TABLE. INSULATED WIRE. 


) 

Gauge 

of 

Wire 

Diameter 
of Wire 
(bare). 

Single Cotton Insulation. 

Double Cotton Insulation. 

Thickness 
of insulation. 
Inch. 

Ratio 

| of bare diameter 
to thickness 
of insulation. 

«i 

C . 1 
O 

O 

S-* 

> t 

Thickness 

of insulation. 

Inch. 

Ratio 

1 of hare diameter 
to thickness 
| of insulation. 

Weight 

1 of insulation 
per 100 lbs. 
of covered wire. 

Weight 

1 of covered wire 
per lb. of bare 
wire, A- g . 

O 

£ 

« 

02 

CQ 

Weight 
of insulati 

I per 100 lb 

| of covered v 

Weight 

of covered 

per lb. of b 

wire, k 6 . 

inch 

min 

1 


.300 

7.62 





.020 

15 

2.28 

1.022 


i 

.289 

7.34 





.020 

14.45 

2.32 

1.023: 

2 

, . 

.284 

7.21 





.020 

14.2 

2.33 

1 023 

3 

■., 

.259 

6.58 





.020 

12.95 

2.40 

1.024- 


2 

.258 

6.55 





.020 

12.9 

2.40 

1.024 

4 


.238 

6.04 





.020 

11.9 

2.50 

1.025 


3 

.229 

5.82 





.020 

11.45 

2.55 

1.025 

5 


.220 

5.59 


9 



.020 

11 

2.65 

1.026. 


4 

.204 

5.18 

.012 

17 

2.20 

1.022 

020 

10.2 

2.85 

1.028 

6 


.203 

5.16 

012 

16.9 

2.20 

1.022 

.020 

10.15 

2.86 

1.028' 


5 

.182 

4.62 

.012 

15.15 

2.27 

1 0227 

.018 

10.1 

2.87 

1.028 

7 


.180 

4.57 

.012 

15 

2.28 

1.0228 

.018 

10 

2:90 

1.029 

8 


.165 

4.19 

.012 

13.75 

2.33 

1 0233 

.018 

9 17 

3.20 

1.032 


6 

.162 

4.12 

.010 

16.2 

2.24 

1 0224 

.018 

9 

3.25 

1.032; 

9 


.148 

3.76 

.010 

14.8 

2.30 

1-023 

.016 

9.25 

3.15 

1.031 


n 

t 

.144 

3.66 

.010 

14.4 

2 32 

1.0232 

.016 

9 

3.25 

1.032. 

10 


.134 

3.40 

.010 

13.4 

2.36 

1.0236 

.016 

8.4 

3.55 

1.035 


8 

.1285 

3.27 

.010 

12.85 

2.40 

1.024 

.016 

8 

3.75 

1.037.' 

11 


.120 

3.05 

.010 

12 

2.50 

1.025 

.016 

7.5 

4.10 

1.041 


9 

.1144 

2.91 

.010 

11.4 

2.55 

1.0255 

.016 

7.1 

4.35 

1 043 

12 


.109 

2.77 

.010 

10.9 

2.66 

1.0266 

.016 

6.8 

;4.60 

1.046 


io 

.102 

2.59 

.010 

10.2 

2.85 

1.0285 

.016 

6.4 

5.00 

1.05 

13 


.095 

2.41 

.010 

9.5 

3.10 

1.031 

.016 

5.9 

5.55 

1.055 


ii 

.091 

2.31 

.010 

9.1 

3.25 

1.0325 

.016 

5.7 

5.85 

1.058 

14 


.083 

2.11 

.007 

12 

2.50 

1.025 

.016 

5.2 

6.60 

1.066 


12 

.081 

2.06 

.007 

11.6 

2.54 

1.0254 

.016 

5.1 

6.80 

1.068 

15 

13 

.072 

1.83 

.007 

10.3 

2.80 

1.028 

.016 

4.5 

7.80 

1.078 

16 


.065 

; .65 

.007 

9.3 

3.15 

1.0315 

.016 

4.1 

8.60 

1.086 


14 

064 

1.63 

.007 

9 1 

3.25 

1.0325 

.016 

4 

8.80 

1.088 

ir 


.058 

1.47 

.007 

8.3 

3.60 

1.036 

.014 

4 1 

8.60 

1.086 


15 

.057 

1.45 

.007 

8.1 

3.70 

1.037 

.014 

4 1 

8.60 

1.086 


16 

.051 

1.30 

.007 

7.3 

4.20 

1.042 

.014 

3.6 

9.60 

1.096 

18 


.049 

1.25 

.007 

7 

4.40 

1.044 

.014 

3.5 

9.80 

1.098 


17 

.045 

1.15 

.005 

9 

3.25 

1.0325 

.012 

3.75 

9.30 

1.093 

19 


.042 

1.07 

.005 

8.4 

3.55 

1.0355 

.012 

3.5 

9.80 

1 098 


18 

.040 

1.02 

.005 

8 

3.75 

1.0375 

.012 

3 33 

10.10 

1 101 


19 

.036 

0.91 

•005 

7.2 

4.80 

1.043 

.005* 

7.2 

5.60 

1.056 

20 


035 

0 89 

.005 

7 

4.40 

1.044 

.005* 

7 

6.00 

1.06 

21 

20 

032 

0.81 

.005 

6.4 

5.00 

1.05 

.005* 

6.4 

6.60 

1.066 

22 

21 

028 

0 71 

.005 

5.6 

6.00 

1 06 

.004* 

7 

6.00 

1.06 

23 

22 

.025 

0.64 

.005 

5 

7.00 

1.07 

.004* 

6.25 

7.00 

1 07 

24 

23 

.022 

0.56 

.005 

4.4 

8.00 

1.08 

.004* 

5.5 

8.00 

1 08 

25 

24 

.020 

0.51 

.005 

4 

8.80 

1.088 

.004* 

5 

8.80 

1.088 

26 

25 

.018 

0.46 

.005 

3.6 

9.60 

1.096 

.004* 

4.5 

9.60 

1 096 

27 

26 

016 

0.41 

.005 

3.2 

10.40 

1.104 

.004* 

4 

10.40 

1.104 

28 

27 

.014 

0.36 

.005 

2.8 

11.25 

1.1125 

.004* 

3.5 

11.25 

1.112 

29 

28 

.013 . 

0.33 

.005 

26 

11.65 

1.1165 

.004* 

3.25 

11.65 

1.116 

30 


.012 

0.31 

.005 

2.4 

12.05 

1.1205 

.004* 

3 

12.05 

1.120 


29 

.011 

• 0.28 

.005 

• 

2.2 

12.45 

1.1245 

.004* 

2.75 

12 45 

1.124 


* Double silk: i mil of silk insulation taken equal in weight to 1.25 mil of cotton 


covering. 
































































46 


DIRECT CURRENT DYNAMOS 



Fig;. 13. 




























































































































































































































































































































































































































































































































DIRECT CURRENT DYNAMOS. 


47 


Calculation of a Bipolar, Single Magnetic Cir= 
cuit, Smooth=Drum, High=Speed Shunt 
Dynamo. 

50 KW. Upright Horseshoe Type. Wrought-Iron 
Cores and Yoke, Cast=Iron Polepieces. 

250 Volts. 200 Amperes. 1050 Revs, per Min. 

(a) Calculation of Armature. 

1. Length of Armature Conductor. 

An electromotive force of one volt is generated by a conduc¬ 
tor that cuts 100,000,000 C. G. S. lines of force per second. As 
the English system of units is still the standard in this country, 
one foot will be taken as the unit length of conductor, one foot 
per second the unit linear velocity, and one magnetic line of force 
per square inch as the unit field strength. Then every foot 
(12 in.) of inductor, moving at the rate of one foot (12 in.) per 
second in a field having one magnetic line of force per square inch 
will generate an E. M. F. of 

12 X 12 X 1 _ 144 

100,000,000 10* 

As the armature of an ordinary bipolar dynamo has two 
parallel conductors each generating the same E. M. F.; and as 
these conductors are in parallel, two feet of conductor will be used 
in generating 144 X 10- 8 volt or in other words each foot of the 
total length of conductors will generate only 72 X 10- 8 volt if 
moving at unit velocity in unit field. 

As this theoretical value of “ unit armature induction ” 
assumes that there is a magnetic field entirely surrounding the 
armature, it will have to be modified so as to take into account the 
fact that the fringe of magnetic lines from the pole pieces only 
partially surround the armature. For a 50 KW. bipolar dynamo 
with smooth-drum armature the polar arc (see figure 14) may be 
taken for the present as about 124°. It is found from actual 
practice that for a machine having a polar arc of about 124°, the 




48 


DIRECT CURRENT DYNAMOS. 


unit armature induction will not be the theoretical value of 
72 X 10-' 8 but will be about 

e — 61 X 10- 8 volts. 

* The “specific armature induction,” i. e., the induction per 
unit length of conductor moving at velocity v c , in a magnetic 
field of strength, 5C," will be 

e' — e X v c X 36" volts. 

where e' = specific induction of active armature conductor, in volts 
per foot of conductor ; 

e — unit armature induction per pair of armature circuits 
in volts per foot of conductor ; 
v c — conductor-velocity, or cutting speed, in feet per second; 

X" = field density, in magnetic lines of force per square 
inch. 

It is customary to take the conductor velocity v 09 as about 50 
feet per second for a 50 KW bipolar dynamo having a drum 
armature; also to take the field density x" as about 22,000 if the 
machine has cast iron polepieces. Therefore the value of e' may 
be written, 


61 X 50 X 22,000 _ 671 
100,000,000 1^00 


Knowing the specific armature induction, e\ the voltage 
induced by one foot of conductor, and knowing the voltage, E\ 
that the armature is required to induce, one may easily find the 
total length of active wire, £ a , of the armature. 




E' _ E X 10 3 
~7 671 


where L a — total length of active conductor (on whole circum¬ 
ference opposite polepieces) ; 

E' ~ total E. M. F. to be generated in armature; i. e ., 
volt output plus additional volts to be allowed for 
drop due to internal resistance. 

For a dynamo of 50 KW. capacity it is necessary to add 
about 6% to the value of E , the voltage wanted by the external 
circuit, in order to get E, the total E. M. F. to be generated in 
the armature. If E is 250, E' is 106% of 250 or 265, and 


E 


265 X 10 3 


395 feet of active conductor. 


671 








DIRECT CURRENT DYNAMOS. 


49 


The part of the conductor passing over the ends of the armature 
core is not active in cutting lines of magnetic force and so this 



SCALE 1 = 1.0 

Fig. 14. Magnetic Circuit of 50 KW, Dynamo. 


length of conductor will have to be increased when the dimensions 
of the iron core of the armature are known. 

2. Sectional Area of Armature Conductor, and 
Selection of Wire. 

-j- The cross section of wire to be chosen should be large enough 
so as not to be unduly heated by the current it has to carry. It 
is well to allow about 600 circular mils of copper conductor to 1 



































50 


DIRECT CURRENT DYNAMOS. 


ampere of current, which is a current density of about 2,100 
amperes per square inch of copper. Therefore to get the number 
of circular mils of copper to be provided multiply 600 by the cur¬ 
rent to be generated. There is to be furnished to the outside 
circuit 200 amperes. The armature will have to furnish this plus 
the current that goes through the shunt field. This latter value 
is so small however that it may be omitted with only trifling error 
or one may choose slightly larger wire for the armature than is 
needed for the 200 amperes. The cross section of conductor 
required therefore will be 

200 X 600 circular mils or 120,000 circular mils. 

As there are two conductors to carry this current each con¬ 
ductor should have 60,000 circular mils. By reference to the 
B. & S. wire table on page 44 it will be seen that No. 2 wire has 
a cross section of 66,371 circular mils. 

As No. 2 wire is too stiff to wind on the armature, however, 



a cable made up of seven strands of No. 11 wire, having a copper 
cross section of 7 X 8,234 circular mils or a total of 57,638 circu¬ 
lar mils is used. This will be a little higher current density than 














DIRECT CURRENT DYNAMOS. 


51 


was calculated upon at first but is not at all excessive. The 
diameter of No. 11 B. & S. wire is .09074 inch and a single cotton 
insulation of .007 inch is put on each strand and the seven strands 
are covered with a double cotton insulation of .016 inch. This 
will give the cable composed of seven strands of No. 11 B. & S. 
insulated wire a diameter (see Fig. 15) of 

S' a = 3 X (.09074 inch -j- .007 inch ) -j- .016 inch 
= .30922 inch. 


3. Diameter of Armature Core. 


* If the speed JYis 1,050 revolutions per minute or ——— per 

second, and d\ is the mean diameter of the armature winding in 
d' 

inches, or_* in feet, then the cutting speed of the conductor in 

12 

feet per second will be represented by, 


d a X 7T v N 
12 60 


or 

jr _ 12 X 60 v v c _ 12 X 60 X 50 _ ln Q1 „ 

a ” 7T A AT 7T X 1050 

It is found from actual practice that the ratio of the diame¬ 
ter of the armature, d^ to the mean diameter of the armature 



scale .r = 1.0" 

Fig. 16. Details of Armature. 


winding, d ' a , is about .95 for drum armatures between eight and 
ten inches in diameter. Therefore (see Fig. 16) 

d a = .95 of 10.91 inches, or 10.36 inches. 












52 


DIRECT CURRENT DYNAMOS. 


4. Length of Armature Core. 

f Knowing the number of conductors that can be laid on the 
surface of the armature core, and the available depth for the 
windings, one may easily determine the number of conductors that 
can be wound on the armature; and knowing this and the length 
of active armature conductor, the length of the armature core may 
be determined. For a machine generating less than 300 volts and 
having a drum armature of the size of the one under consideration 
about 8% of the surface of the armature core is given up to divi¬ 
sion strips or driving horns. 


* Then — 


.92 X d & X tt 

V 

.92 X 10.36 X tt 
.30922 


9T or 96. 


where n w = the number of armature wires per layer; 

<4 = the diameter of the armature core, in inches; 

S' a z= the width of insulated armature conductor, in inches; 
.92 = the portion of the surface to be occupied by the 
conductors. 


For a drum armature of the size of the one under considera¬ 
tion the height of the windings should be about .55 of an inch 
and of this space about .06 inch will be taken up by the insula¬ 
tion between windings and the armature, and .05 inch by binding 
wires. This leaves a net height of about .14 of an inch for the 
conductor. Then the number of layers of conductors will be 


Til 



.44 

.30922 


= li or 2 


where n x = number layers of armature wire; 

— net height of winding space, in inches; 

— height of insulated conductor, in inches. 


By dividing the length of active armature conductor by the 
number of conductors we obtain the length of one active conduc¬ 
tor which is the length of the armature core (see Fig. 16). 


_ 12 X L a 
X n Y 


12 X 395 
96 X 2 


= 24| inches approx. 


where Z a == length of armature core parallel to pole faces, in inches; 
Z a = length of active armature conductor in feet; 
n w = number of wires per layer; 
n x — number of layers of wire on armature. 








DIRECT CURRENT DYNAMOS. 


53 


5. Arrangement of Armature Windings. 

f For machines under 300 volts it is customary to have from 
40 to 60 commutator segments. With this number of segments 
the voltage between two consecutive ones is low, and the pulsating 
current is within a fraction of one per cent of being a steady cur¬ 
rent. The number of commutator divisions, n c , will be found by 
multiplying the number of wires per layer, w w , by the number of 
layers, %, and dividing the product by some even number that 
gives a quotient which is between 40 and 60. 

n c = n w X % -f- 4 = 48. 

The number' of convolutions per commutator segment, will 
be 

n _ X n x __ 96 X 2 _ 

a 2 X n c 2 X 48 

since it takes two conductors to make one turn. Therefore to sum 
up we have 48 coils, each consisting of 2 turns of a cable having 

7 No. 11 B. & S. wires. 

6. Total Length of Armature Conductor, Weight 
and Resistance. 

In order to connect the ends of the active conductors, turn 
the corners, etc., for this drum armature the total length of arma¬ 
ture conductor will need to be about 1-| times the active conductor. 
L t — 1.T5 X A a = 1.75 X 395 feet = 691 feet. 

The weight of the conductor will be as follows: — A copper 
wire .001 in diameter weighs .00000303 pounds per foot of length. 
Therefore the weight of the total length of the copper in the 
armature conductor would be 

wt a = & 5 X A X 3 a 2 X .00000303 

= 1.03 X 691 X 57638 X .00000.303 = 124 pounds* 
where wt & — weight of bare armature winding in pounds ; 

Jc 5 — ratio between weights of the insulated wire and bare 

wire; 

L t = total length of armature conductor in feet; 

8 a 2 •= area of conductor in circular mils. 

The resistance of the armature will be as follows : — The 
total length of armature wire 691 feet is arranged in two parallel 





54 


DIRECT CURRENT DYNAMOS. 


circuits 3451- feet long. These two paths are each composed of 
seven No. 11 B. & S. wires. 

The resistance of 1,000 feet of No. 11 B. & S. wire is 1.311 
ohms. 

Therefore the resistance of the armature is, 

1 qjs 5 

•Ba = 2“xT X T(M X 1-311 ohmS ~ 0324 ohm ‘ 


7. Radial Depth of Armature Core, Minimum and 
Maximum Cross Section, and Average Mag¬ 
netic Density of Armature Core. 

*The following formula determines the proper size for the 
necessary strength of the armature shaft where it passes through 
the iron core of the armature. 


d c — X 


= - X 


000 


050 


=■ 3.42 inches. 



where d c — diameter of armature shaft at core 
in inches ; 

P' — capacity of machine in watts; 

N = speed, in revolutions per minute; 
Jc 9 — constant depending upon capa¬ 
city of machine. 

Therefore the breadth of armature cross 
section, or radial depth of armature core is 


(<*» — <*c) 


10.36 — 3.42 


— 3.47 inches. 


where 5 a = radial depth of armature core, in inches; 
d a = diameter of armature core, in inches. 
d c = diameter of core section of armature shaft. 

The maximum depth of armature core (see Fig. IT) is, 



= 4.89 inches. 

















DIRECT CURRENT DYNAMOS. 


55 


f The cross section of the magnetic field in the armature is 
equal to 

= 2 X Z a X X K 

— 2 X 24.75 X 3.47 X -95 = 163 square inches. 

= 2 x ?a x v % X & 2 

= 2 X 24.75 X 4.89 X -95 = 230 square inches, 
where S"^ = minimum cross section of armature core, in square 
inches; 

N" a2 = maximum cross section of armature core, in square 
inches; 

Z a = length of armature core, in inches; 
b r A = radial depth of armature core, in inches ; 
fc 2 ~ ratio of net iron section to total cross section of 
armature core. 

The useful magnetic flux in the armature may be obtained as 
follows : The 

t~> if Number of C. G. S. lines cut per second 

E. M. F. = -iol---- 


v __ Jsr x iv c x <e> 


volts. 


60 X 10 8 

where E' — total E. M. F. induced in armature; 

N — number of armature revolutions per minute ; 
iV c = total number of conductors all around pole facing sur¬ 
face of armature; 

<I> = total number of useful magnetic lines, in webers. 

60 X 10 8 X E' 


or <I> = 


NX N c 

60 X 10 8 X 265 


— 7,886,905 webers. 


1050 X 192 

Therefore the density of magnetic lines per square inch at the 
minimum cross section of the armature core is 
„ 7,886,905 


al 


163 


and the density at the maximum cross section is 
7,886,905 


®"a 2 = 


230 


= 48,386 lines per square inch, 
imum cross section is 
— 34,291 lines per square inch. 










56 


DIRECT CURRENT DYNAMOS. 


Next we wish to find the number of ampere-turns required 
to furnish this induction. The law for the magnetism in a mag¬ 
netic circuit is exactly similar to the law for current in an electric 
circuit. 

Electromotive Force 
Resistance 

A/r T-yi Magnetomotive Force 

Magnetic blux = -5- 

Reluctance 

Therefore the 

Magnetomotive Force = Magnetic Flux X Reluctance. 

As the resistance of an electric circuit can be expressed by 
the specific resistance or resistivity of the material multiplied by 
the length of circuit and divided by the cross section, so the 
reluctance of a magnetic circuit can be expressed by the specific 
reluctance or reluctivity of the material multiplied by the length 
of the circuit and divided by the cross section. Therefore the 

Reluctance — Reluctivity X . 

Area 


Current (or Electric Flux) 


As the conductivity of an electric circuit is the reciprocal of 
the resistivity, so the permeability of a magnetic circuit is the 
reciprocal of the reluctivity. Therefore the 

-d i x Length 

Reluctance =-___ 

Permeability X Area 


or 


Magnetomotive Force = Magnetic Flux X Length' 

Permeability X Area 

and since the magnetic flux divided by the area is the magnetic 
density the formula is simplified by writing 

Magnetomotive Force = Magnetic Density X Length 

Permeability 

The Unit of Field Density, or the magnetic density caused by 
a unit pole, is 1 line of magnetic force per square centimeter of 
field area and is termed 1 gauss. 

A single Line of Force, or the Unit of Magnetic Flux, is that 
amount of magnetism that passes through every square centi¬ 
meter of cross section of a magnetic field whose density is unity, 
and is termed 1 weber. 









DIRECT CURRENT DYNAMOS. 


57 


The unit magnetic pole, or the pole of unit strength is that 
which repels an equal pole at unit distance with unit force. The 
lines of force are straight lines from the center of the sphere to 
the surface, there being one line to each square centimeter area 
on the surface. As the surface of a sphere having a radius of 1 
centimeter has an area of 47r square centimeters, it follows that 
from a pole of unit strength there is a magnetic flux of 4 tt C. G. S. 
lines of magnetic force or 47 t webers or 12.5664 webers. 

One absolute C. G. S. unit of current, which is 10 times as 
large as the ampere, or 10 amperes, flowing in a wire 1 centimeter 
long bent into an arc of a circle of 1 centimeter radius gives a 
C. G. S. unit magnetic pole at the center of curvature or 12.5664 
webers. One practical unit or 1 ampere would cause one tenth as 
many webers or 1.25664 webers. 

A long solenoid having a cross section of one square centi¬ 
meter having 1 ampere (A^ of the C. G. S. unit current) flowing 
per unit length of coil, has poles of A^- unit strength which causes 
a magnetic flux of A^ X 4 77 - webers. 

• • 47T 

The density of the magnetic circuit is-webers per square 


centimeter or 


47T 

TO 


gausses. 


The reluctance of unit length of the solenoid of one square 
centimeter cross section for air is unity or 1 oersted. The mag¬ 
netomotive force is the product of the magnetic density, the 
reluctance and the length, and is measured in gilberts. Therefore 
the magnetomotive force required to produce a magnetic density 


of 


47T 

TO 


gausses in a column of air one centimeter long and having 


a cross section of one square centimeter thus having a reluctance 
of 1 oersted is 


M. M. F. = ^ X 1 X 1 — gi^erts. 

The magnetomotive force of gilberts being produced by 


one ampere-turn, it follows that the 

Number of Ampere-turns = X Number of Gilberts. 

47T 



58 


DIRECT CURRENT DYNAMOS. 


The 

Magnetizing Force = Specific Magnetizing Force X Length, 
or the 

Number of Ampere-turns = Ampere-turns per unit of 
Length X Length, 

or at—f (©") X l. 

where at — Ampere-turns required to magnetize a portion of a 
magnetic circuit; 

(&" = density of the magnetic circuit per square inch; 

/ (&") = Specific magnetizing force, in ampere-turns per inch 
of length for the particular material and density 
employed; (this value of the magnetizing force 
must be taken from some table or induction curve 
as shown in Fig. 13, or found by experiment for 
the particular piece of iron to be used) ; 
l i= length of magnetic circuit of the material in inches. 
To get the number of ampere-turns required to overcome the 
reluctance of the armature it is necessary to modify the above 
formula a little, as the value of ($>" is not constant in all parts of 
the core. 

/ (<B'0 = \ \f («"*) +/ («'u > j 

_/ (48,386) +/ (34,291) _ 9.2 + 6.4 _ 7 a 
2 2 

It takes as seen from Fig. 13 about 9.2 ampere-turns to mag¬ 
netize 1 inch of wrought iron to a density of 48,386 lines, and 6.4 
ampere-turns to magnetize it to a density of 34,291 lines. Also 
7.8 ampere-turns correspond to a density of 41,500 lines. 

8. Energy Losses in Armature, and Temperature 
Increase. 

The energy lost in the armature due to the current in the 
armature conductors is 

P a = 1.2 X (*, X iy X r a 

= 1.2 X (1.03 X 200)2 x . 03 24 = i 65 0 watts. 





DIRECT CURRENT DYNAMOS. 


59 


where P a = energy dissipated in armature winding, in watts. 
r a = resistance of armature winding cold, in ohms ; 

1.2 = ratio of resistance of armature winding hot, to resist¬ 
ance cold; 

I — output of machine in amperes ; 

1.03 = ratio of total current generated to output. 

The resistance of copper wire at 150° F. is about 1.2 times 
that at 60° F. The energy in the shunt coil of a 50 KW. com¬ 
pound dynamo is about 3 % of the output of the machine. 

The loss in the armature core due to hysteresis is proportional 
to the 1.6th power of the magnetic density, directly proportional 
to the number of magnetic reversals, and directly proportional to 
the mass of the iron. Expressed in C. G. S. absolute units the 
energy consumed by hysteresis is, 


P h r = n l X « a 16 X X M 1 , 

iY = wj x (B a 16 X JST X X 

where P' h = energy lost due to hysteresis, in ergs; 

n 1 = constant depending on magnetic hardness of material 
“ Hysteresis Resistance; 

® a = density of lines per square centimeter of iron; 

N’ 1 — frequency, or number of complete cyles of 2 revers¬ 
als each, per second; 

M x ' — mass of iron in cubic centimeters. 

For soft sheet iron discs n 1 may be taken as .0035. In order 
to get the value of P h ' in practical units we must change the 
above equation. One watt equals 10 7 ergs. Instead of © a use 
the density <£ a " in lines per square inch. 


^i = 


N_ 

60 


l 050 - = 171. 
60 2 


One cubic foot equals 28,316 cubic centimeters. The mass 
of iron in the armature in cubic feet is 


M = 


dj " X 7r x x X lc 2 
~~ 1,728 


_ (10.36 - 3.4T) X tt X 3.47 X 24.75 X .95 = 1>02 cubic 
“ 1728 






60 


DIRECT CURRENT DYNAMOS. 


where M = net mass of iron, in cubic feet; 

dj" = mean diameter of armature core, in inches, 

= <4 — K, ( see Fi g- 17 ) 

Z a = length of armature core, in inches ; 

b a = radial depth of armature core, in inches; 
k 2 = ratio of net iron section to total iron section. 

There are 1,728 cubic inches in 1 cubic foot. Expressing 
the value of the energy lost by hysteresis in practical units we 
have, 

P h = 10-7 x .0035 x x 28,316 X X M, 

where P h = energy lost by hysteresis, in watts; 

CB a " = density, in lines per square inch, corresponding to 
average magnetizing force required for amature 
core, (1 square inch = 6.45 square centimeters) ; 
N x = frequency, in cycles per second; 

M — net mass of iron in amature, in cubic feet. 

By reducing the above expression to simple form, we have 
P h = 5 X 10-7 x (B a " 16 X iV, X M, 

= 5 X 10-7 X (41,500) X 171 X 1.02= 219 watts. 

Eddy Currents. 

It has been shown under Electromagnetic Induction (see 
theory of Dynamo-Electric Machinery , p. 4 etseq.) by what means 
and methods electromotive forces, and consequently electrical cur¬ 
rents are set up in closed conductors. It is clear that any mass of 
metal moving in a field is a closed conductor. A loop of wire may 
have currents generated in it quite as easily if it be made part of 
a metallic disc as if it were still a loop. Since the lines of force 
can only cut a solid piece of metal once, the electromotive force 
that will be generated, unlike that in a coil of many turns cut by 
the same field will be very small. But if the piece of metal be 
large the resistance will be very small so that the current which is 
induced may reach considerable strength and cause much heating. 
This action is largely prevented from taking place in the arma¬ 
ture cores of dynamos by building them up of thin sheets of iron. 
The sheets are insulated from each other and are placed in such a 




DIRECT CURRENT DYNAMOS. 


61 


direction as to cut across the path which would be followed by the 
induced current. This method of construction is termed lamina¬ 
tion, and it is necessary to build all iron or metallic parts which 
are not intended to act as conductors, in this way, if they are 
likely to be subjected to fields of varying strength. This is the 
case in alternating current machinery and in the moving parts of 
direct current machinery. Sometimes even conductors if they are 
very large, have local currents generated in them which are of 
course undesirable because they are a source of waste and heat, 
and it becomes necessary to laminate them. In this case the 
lamination would be parallel to the length of the conductor. 
Heavy conductors are usually laminated or stranded any way, for 
greater ease in construction. Eddy currents are often called 
Foucault currents from Foucault who first called attention to their 
existence. 

The energy lost by eddy currents is found to be proportional 
to the square of the magnetic density, to the square of the fre¬ 
quency, and to the mass. The equation for the lost energy due 
to eddy currents is, 

PJ = € ' X (B a 2 X iV X M\ ; 

where P e ' = lost energy due to eddy currents, in ergs; 

(B a = density of lines of force, per square centimeter of 
iron; 

= frequency, in cycles per second ; 

M\ r = mass of iron, in cubic centimeters ; 

e' = eddy current constant, depending upon the thickness 
and the specific electric conductivity of the 
material ; 

7T 2 

e'= -g- X S 2 X 7 X 10-9 

= 1.645 X 8 ! X 7 X 10-9 
where 8 = thickness of sheet iron, in centimeters ; 

7 = electrical conductivity, in mhos; 

7 = 100,000 for iron. 

By changing to practical units and simplifying the above 
equation, 



62 


DIRECT CURRENT DYNAMOS. 


-P. = 10-T X 1.645 X (2.54 8,)* X 1<M X j T ®U J * X JVj* 

X 28,316 X M= 7.22 x 10- 8 X X <B a " 2 X X M. 
where = thickness of iron laminse in armature, in fractions 
of an inch; 

($>J f = density, in lines per square inch, corresponding to 
average specific magnetizing force of armature 
core; 

iVj == frequenc}^ in cycles per second ; 

M = mass of iron, in cubic feet. 

The terms 7.22, 10- 8 , and (B a " 2 , may be multiplied 
together to form the eddy current factor e, and then the formula 
for loss due to eddy currents becomes, 

P e = e X 2 X M 

= .0125 X (17J) 2 X 1«02 = 4. watts. 

For e = .0125 when the thickness of the sheet iron discs is 
.01" and when the magnetic density is 41,500 lines. 

The total energy loss in the armature is 

P A = P a, ~h P\i + Pe = 1650 -f- 219 + 4 

= 1873 watts = 2.5 h. p. 

Knowing the amount of electrical energy lost in the armature 
and dissipated as heat, and knowing the dimensions and speed 
of the armature it is possible to calculate quite exactly, by using 
certain known constants that have been found by experience, 
to calculate the rise of temperature in the armature and to 
calculate the hot resistance of the copper in the armature con¬ 
ductors. These details are somewhat complicated and will not be 
treated. 


(b) Dimensions of flagnet Frame. 

1. Total Magnetic Flux, and Sectional Area of 
Magnet Frame. 

The total magnetic flux to be generated in a dynamo is the 
useful magnetic flux multiplied by a factor of magnetic leakage. 
Knowing the shape of the various parts of the magnetic circuit and 
the reluctance of the parts of the magnetic circuit and the sur- 




DIRECT CURRENT DYNAMOS. 


63 


rounding air circuits, it is possible to determine the amount of 
magnetic leakage, or waste magnetism. 

<$>' = X X <S> = 1.35 X 7,886,905 = 10,647,322 webers. 
where = total flux to be generated in dynamo, in lines of force ; 

<E> — useful flux necessary to produce the required E. M. F. 

X = factor of magnetic leakage, which for a 50 KW. 

bipolar machine of the Edison type is found to be 
about 1.35. 

The sectional area of the magnet frame is expressed by the 
formula, 


& _ 10,647,322 
^ 90,000 


118 square inches. 


where — cross section of wrought iron magnet core and of 
yoke, in square inches ; 

— total flux, in webers ; 

= magnetic density of magnet frame, which for wrought 
iron magnet cores and yoke is taken as 90,000 
lines per square inch. 

The Practical Limit of Magnetization for cast iron pole pieces 
is taken as 50,000 lines per square inch. 

_ 10,647,322 
50,000 


sr 


= 213 square inches. 


2. Sectional Area of flagnet Frame. 


If the magnet core is in the form of a cylinder the diameter 
will be 



118 = 12.25 inches. 


For a circular magnet core carrying about 10,000,000 lines of 
force the most economical ratio of length to diameter of magnet 
core is found to be about 1.5. Hence 

Z m = 1.5 X = 18.375 inches. 

For a machine having an armature between 10 and 11 inches 
in diameter the distance, c, between magnet cores is taken as 
about 

c — 5-| inches. 


Take the width of the wrought iron yoke a little greater than 






64 


DIRECT CURRENT DYNAMOS. 


the diameter of the magnet cores so as to form a mechanical pro¬ 
tection to them. Let this width be 18J- inches. Then the height 
of the yoke would be 

h y = = 64 inches. 

y 18J 1 

The polepieces are determined as follows : — The bore of the 
polepieces is the sum of the diameter of the armature core, the 
winding, the insulation and binding, and the air gaps, which for a 
drum armature of this diameter is taken as about .14". 


Diam. of bore = 10.86" + 4 X .30922" + 2 X .11" + .14" 
= 11.96". 

In order to keep the corners of the polepieces far enough 
apart so that there will not be too much leakage across, the dis¬ 
tance is taken from 1.25 to 8 times the length of the two air 
gaps according to the size and type of dynamo. 


V v = Jc n X (d v — c? a ) = 4. X (11-86 — 10.36) = 6.00 
inches. 

where l' v = distance between pole corners, in inches; 
d v = diameter of bore of polepieces, in inches; 

4 = diameter of iron core of armature, in inches ; 
k xl = 4. for drum armature of a 50 KYV. bipolar dynamo. 
Let the length of the pole pieces equal the length of iron 
core in armature, or 

/ _ 943" 

°v - ^4 • 

Let the height of polepieces equal the diameter of bore approx. 


V = 12 "- 


The thickness at the center which carries only half the lines 
will be 


213 

2 X 24| 


4.3 + or 4f". 


In order to provide a non-magnetic 'base through which the 
magnetic lines will not leak from pole to pole a block of zinc about 
5 inches thick should be placed under the dynamo. 




DIRECT CURRENT DYNAMOS. 


65 


(c) Calculation of Magnetizing Forces. 


1. Air Gaps. 

The length of the two air gaps is 
l\ = Jc 12 X (d v — O = 1.4 X 1-5" = 2.1" 
where Z" g = length of path of magnetic lines across the two air 
gaps; 

d v = diameter of bore of pole pieces, in inches ; 
a — diameter of iron core of armature, in inches ; 

Jc 12 = constant depending on the path of the lines of force 
through the air gap. The lines of force pass through the gaps 
obliquely owing to the distortion of the magnetic field and the 
constant, & 12 , grows greater as the velocity of the conductor, v Ct> 
and the density, X, are greater. If the product of v c and X is 
above 2,000,000 the value of k 12 is taken as 1.4. 

The cross section of the magnetic field of the air gap is repre¬ 
sented by, 

St = d t X | X ft X l f 

— 11.11" X — X -84 X 24-| 363 square inches. 


where S f = the area occupied by the effective conductors, in square 
inches; 

d t — mean diameter of magnetic field in inches ; 

= i (<4 + ^p) ; 

l { = breadth of magnetic field, in inches ; 

= ratio of effective field circumference, depending upon 
the percentage of polar embrace, here taken as .84. 
The actual field density in air gaps is found by dividing the 
value of the total useful flux in webers, by the area of the 
magnetic field of air gap. 

d> _ 7,886,905 _ 

363 


X" = 


21,727 lines per square inch. 


The number of ampere-turns required to produce this mag¬ 
netic density in the air gaps is* 

l", 


a t K — 12. X X' 
* 4tt 


X 


= .3133 X X" X l\ 
2.54 * 


= .3133 X 21,727 X 2.1 = 14,295 ampere-turns. 





66 


DIRECT CURRENT DYNAMOS. 


2. Armature Core. 

The average length of the magnetic paths in the armature is 
(see Fig. 18). 

l\ =d'\X ttX 

= 6.89" X 7T X ^ + 8.47" = 10,57 inches. 

8 o0 

where Z" a = length of magnetic path in armature core, in inches; 
d"\ == mean diameter of armature core, in inches; 

Z> a = radial depth of armature core, in inches; 
a — half angle between adjacent pole corners. 



Fig. 18. Magnetic Details of Armature Core. 


As previously determined the minimum cross section of the 
armature core is, 

$" al = 168 square inches, 
and the maximum cross section is 
S" 2 & = 280 square inches. 

The average specific magnetizing force to magnetize the arma¬ 
ture core to the required density will be 

/(«'.) = / (48,386) +/ (34,291) = 7>g ampere . turnS- 

Jj 





























DIRECT CURRENT DYNAMOS. 


67 


The total magnetizing force required to magnetize the arma¬ 
ture is 

a £ a = 10.57 X 7.8 = 82 ampere-turns. 


3. Wrought Iron Field Cores and Yoke. 

The length of the magnetic path through the wrought iron 
cores and the yoke will be (see Fig. 14), 

V\ x . = 2 X 18f + 6J + 18 =61 inches, (approx.) 

The density of the field has already been decided upon as 
90,000 lines per square inch. It has been determined by experi¬ 
ment that a specific density of 90,000 lines in wrought iron is 
produced by 50.7 ampere turns. 

The total length of magnetic circuit through field cores and 
yoke will require 

a t w it = 61 X 50.7 = 3093 ampere-turns. 


4. Cast Iron Pole Pieces. 


The average length of the magnetic path through the pole 
pieces may be taken as the average of the longest path and the 
shortest path. In the present case it is about 
l" c i — 15 inches. 

The minimum cross section at the center of the pole piece 
which carries only half the lines has already been taken as 
S C ' iwl = 106.5 square inches 
which corresponds to a density of 

(&" ci — 50,000 lines per square inch. 

The maximum cross section of the magnetic circuit in pole 
pieces is the area of the pole face and is 


'c.i.2 


1 94 

11.86 XirX „^-X 24.75 




which corresponds to a minimum density of 

„ 7,886,905 


c.i .2 


318" 


24802 magnetic lines. 

The average specific magnetizing force will be 

/ = ft/ ( 50 ’ 000 ) +/(2^.802)] = 160 + --- 

= 99.3 ampere-turns, 

which corresponds to average density of 42,175 lines. 





68 


DIRECT CURRENT DYNAMOS. 


Then the total number of ampere-turns required to magnet¬ 
ize pole pieces is 

a t cL = 15 X 99.3 = 1490 ampere turns. 


5. Armature Reactions. 


The demagnetizing and cross-magnetizing effects caused by 
the ampere-turns of the armature oppose the magnetizing effect 
of the field coils and have to be overcome by adding more ampere- 
turns to those required to overcome the reluctance of the circuit. 

* The number of ampere-turns required to balance the arma¬ 
ture reaction is, 


a t T = & 14 X 


^ X I 


X 


13 


X a 


180 c 


= 1.71 X 


96 X 200 


X 


28 


180 


=2554 ampere=turns. 


where a t r = ampere-turns required to compensate for armature 
reactions; 

Tc 14 = being a constant depending upon the magnetic den¬ 
sity in pole pieces; 

jY a = total number of turns on armature; 

1 = total armature current in amperes. 

2 = number of armature circuits for current; 

h 13 X cl — angle of brush lead, which is nearly equal to half 
the angle between two pole corners, for smooth 
drum armatures. 

f The total number of ampere-turns required on field coils of 
dynamo will be the sum of all the ampere-turns required for the 
various parts of the circuit. 

AT — atg -{- at a -f- cit w ^ -j— at c j -[- at T 

= 14,295 + 82 + 3093 + 1490 + 2554 

= 21,514 ampere=turns. 


(d) Calculation of Magnet Winding. 

The shunt winding should be calculated for a temperature 
increase of about 15°C above the normal temperature. (If desired 







DIRECT CURRENT DYNAMOS. 


69 


the shunt coil may be calculated so as to give a higher voltage to 
the dynamo which can be decreased by a regulating resistance in 
series with the shunt coil. As this calculation is not required in 
order to show the electromagnetic theory of the dynamo it will not 
be discussed here. It requires a recalculation of the magnetic flux 
for the various parts of the magnetic circuit and a recalculation of 
the corresponding ampere-turns.) 

1. riagnet Winding. 

The mean length of one turn of wire on the- field core is 
l T = Jc 17 X = 3.66 X 121" = 44.8" 

where k 17 = a constant, depending upon the size of the field 
core, giving the ratio of the length of a mean 
turn to the core diameter. 


The specific length of magnet shunt wire in feet per ohm is 
given by the formula, 

a _ -^sh 

A-sh — - 

^sh 

= -Af X A- X (1 + -004 X 0 m ) 


= 21 ’ 514 X X (1 


.004 X 15) 


250 12 

= 840.5 feet per ohm. 

No. 15 B & S has 315 feet per ohm. 

No. 14 B & S has 39T feet per ohm. 

Use about No. 14 B & S wire. 

The height of the winding space is given by the formula 
L , 44.8 


- ''T _ 

- —— u r 

IT 


12.25 = 2 inches. 


The radiating surface of the magnet coils will be 
= (12.25 + 2 X 2) 7 T X 2 (18.375) = 1877 square inches. 







70 


DIRECT CURRENT DYNAMOS. 


The energy absorbed in the magnet winding may be ex¬ 
pressed by the formula, 


sh 


75 


X -S', 


= -^ X 1877 - 375 watts. 

75 

where P sh = watts absorbed in field winding; 
6 m rise of temnerature in map-nets 


watts absorbed in held winding; 
rise of temperature in magnets, in degrees Centi¬ 
grade ; 

coils. 


grade 

$ M = radiating surface of magnet 
^sh — -gr or ^sh E — P sh . 


Therefore the number of shunt turns may be found by 
7VT - A T - ATX E 

- XY Sh- “r - - --^- 


i.h 


sh 


21514 X 250 , . ... . + + 

= -——- = 14,343 shunt turns, 

875 


is 


375 

The length of the shunt winding 

T __ 14,343 X 44.8 _ , . 

Ish — --- — 53,547 feet, 

1 A 

The resistance of the shunt wire will be 


^sh- o 


53,547 


■ - — 157 ohms, at 15.5° C. 

340.5 

The warm resistance at 30.5° C is 
r sh = 157 X (1 4- -004 X 7 5) = 166 ohms. 

The shunt current at full load is 
T 250 

4h = TTjTT = 1-5 amperes. 

166 

If desired it is a simple matter to get the weight of the wire 







DIRECT CURRENT DYNAMOS. 


71 


(e) Calculation of Efficiency. 

The electrical efficiency of the dynamo is the ratio of the 
available energy to the available energy plus the energy lost 
due to armature resistance plus that due to field coil resistance. 

_ P 

Ve p + p m + p x 

_ 250 X 200 

250 X 200 + 1.06 X 201.5 2 X .0324 + 1.52 x 166 

= .96 or 96$ electrical efficiency. 

The commercial efficiency is the ratio of the output, to the 
sum of the output, wire loss in armature, wire loss in field, hysteresis 
loss, eddy current loss, and friction loss (which we will assume as 
2,500 watts). 

__ P 

Vc ~ p + p a + p M + 1 \ + p e + 

50,000 

“ 50,000 + 1,894 + 374 + 219 + 4 + 2,500 

= .917 or 91.7$, commercial efficiency. 


































































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A • « .'l 

— 

































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M , - 





















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■ 














■ 















DESIGN OF 


Direct Current Dynamos. 


EXAMINATION PAPER 

American 

School of Correspondence 

Copyrighted 1898 

BY 

American School of Correspondence 


BOSTON, MASS., 
U. S. A. 




































. 
































. 


. 


























■ 



















. • ' 








































■ 














■ 


















. 














DESIGN OF 


DIRECT CURRENT DYNAMOS. 


Instructions to the Student. Place your name and full address at the 
head of the paper. Work out in full the examples and problems, showing each 
step in the work. Mark your answers plainly “ Ans.” Avoid crowding your 
work as it leads to errors and shows bad taste. Any cheap, light paper like the 
sample previously sent you may be used. After completing the work add and 
sign the following statement. 

1 hereby certify that the above work is entirely my own. 

(Signed) 


1. (a) What is the value of 9 horse-power expressed in 
watts? ( b ) In kilowatts? (1000 watts = 1 kilowatt). 

2. ( a ) What is the value of 2400 watts expressed in horse¬ 
power? (5) What is the equivalent value of a 10 kilowatt 
dynamo expressed in horse-power ? 

3. (a) If a Phoenix arc light dynamo has an armature 

resistance of 3.448 ohms, and a series coil resistance of 4.541 
ohms, what will be the lost voltage in the machine when a current 
of 10 amperes is flowing? ( b ) What will be the lost voltage 
when a current of 9.8 amperes is flowing ? ( c ) What amount of 

energy in watts is lost in the armature for both of the above 
cases ? ( d ) What amount is lost in the series field? (e) Express 
the total loss in horse-power in each case. 

4. (&) If in a closed circuit the resistance between two 
points is 4.6 ohms what current must flow to cause a difference of 
potential of 20 volts? ( b ) If this current flows for 24 hours how 
many coulombs of electricity will have passed (from table page 
4,1 coulomb == quantity conveyed in 1 second by 1 ampere). ( c ) 
Also what energy in kilowatt-hours will be expended in the 
same time (1 kilowatt-hour equals the energy expended in one 
hour when the activity is one kilowatt). 




76 


DIRECT CURRENT DYNAMOS. 


5. (a) If a current of 100 amperes is transmitted a distance 

of 200 feet over No. 1 B. & S. copper wire, for how many volts 
should the dynamo be overcompounded in order to make good the 
lost voltage due to line resistance ? (Refer to wire table, and 
note that to transmit a current 200 feet requires 400 feet of wire). 
(6) How much energy is lost in the line ? - 


. < (a) 5 volts approx. 

11S * C (&) 500 volts approx. 


6. (a) With 10 amperes flowing, with resistances as stated 
in question 3, and with R = 300 ohms what is the electrical 
efficiency of the dynamo ? ( b ) What is the total horse-power 

generated by the armature? (c) What is the useful output of 
the machine in horse-power? 


7. (a) If an arc lamp gives 2000 candle power at 50 volts 
and 10 amperes, how many candle power is furnished per horse¬ 
power in the case of arc lamps? ( b ) If a 16 candle power incan¬ 
descent lamp requires 1.2 amperes at 50 volts how many candle 
power is furnished per horse-power in the case of incandescent 
lighting ? 

8. The* resistance of three branches of an incandescent 
lighting circuit with all the lamps burning are 5 ohms, 16 ohms, 
and 12 ohms, what is their joint resistance, the three being in 
parallel ? 


9. Explain the reasons for over-compounding a dynamo. 

10. Having the internal and external characteristic curves 
of a dynamo, how would you calculate the electrical efficiency ? 


11. (a) Give total resistance of circuit in which 12.5 

amperes is flowing, the total E. M. F. developed being 100 volts? 
(5) If the external resistance is 10 times the internal, what is the 
external resistance in ohms ? (c) The internal resistance ? 

QcT) What is the useful output of the dynamo ? ( e ) What is the 

electrical efficiency ? 

12. If the resistance of the series coils of a dynamo is 10 
ohms at 15° C. what will be the resistance after the temperature 
1 as risen to 35° C. ? 



DIRECT CURRENT DYNAMOS. 


77 


18. (a) Explain the effect of the shunt coils of a compound 

dynamo. (5) Explain the effect of the series coils. 

14. Explain quite fully the differences between a constant 
current dynamo and a constant potential dynamo. 

15. ( a ) The total electromotive force of a series dynamo is 

1000 volts, and the resistance of the armature and series coils is 
40 ohms. If a current of 10 amperes is flowing what will be the 
resistance of the external circuit ? (5) What will be the elec¬ 
trical efficiency of the dynamo ? ( [c ) What will be the total 

horse-power generated by the armature ? 

16. (a) If the external E. M. F. of a dynamo is 550 volts 
and 10% of this voltage is lost in transmitting a current of 580 
amperes a distance of one-half mile (one mile of wire required) 
what is the resistance of the conductors? ( b) What is the 
resistance per foot of conductor? (c) About what size of copper 
wire would be required ? 

17. (a) If a ring-wound or drum-wound armature of a 

bipolar ordinary compound-wound dynamo has 200 complete turns 
of copper conductor, if the total magnetic flux cut by these con¬ 
ductors is 12,000,000 magnetic lines of force, and if the armature 
runs at 1500 revolutions per minute, what is the total E. M. F. 
in volts generated in the armature ? (5) If at full load the lost 

voltage in the armature is 2J% and the value of r sh is 500 
ohms, what current flows through the shunt field coils ? (c) If 
2J% of the energy generated is lost in the armature, 2% in the 
field coils, and \\% in the series coils, what is the electrical 
efficiency of the dynamo ? (c?) What is the total output in kilo¬ 

watts ? 

18. (a) In the following equation explain the meaning of 
each of the symbols used 

_ 1_ 

1 4- R ( rsh T -i- 4- 9! r& 

r 2 sh ^ R ^ r* 

( l ) For what type of dynamo does this equation give the electri¬ 
cal efficiency? 





78 


DIRECT CURRENT DYNAMOS. 


19. (&) By wliat means can the electrical efficiency of a 
dynamo be increased? (5) In order that a shunt dynamo shall 

have an efficiency of 95% what is the value of rsh - ? (c) Are 

there any objections to having the value of as great as 100,- 

^*a 

000 for small dynamos? (d') If the electrical efficiency of a 5000 
horse-power dynamo is 98% how much electrical energy is lost, 
expresed in kilowatts. 

20. ( a ) Explain the meaning of the Electrical Efficiency of a 

dynamo. (5) Explain the meaning of the Commercial Efficiency 
of a dynamo. (c) What is a Saturation Curve? (c?) What 

is an External Characteristic? ( e ) Internal Characteristic? 

(/) How would you construct a horse-power curve ? 

£21. Draw as nearly as you can the shape of the internal 
and external characteristic curves of a shunt dynamo of about 5 
horse-power, beginning with an initial E. M. F. of 115 volts. 
Draw the necessary horse-power curves. (Make ordinates 20 
volts per inch, and abscissae 10 amperes per inch.) 

22. Draw as nearly as you can internal and external charac¬ 
teristic curves for a series dynamo, the maximum energy to be 
about 10 horse-power. (Make ordinates 100 volts per inch, and 
abscissae 5 amperes per inch.) 

28. Draw as nearly as you can internal and external charac¬ 
teristic curves of a compound dynamo, the maximum energy to be 
about 8 horse-power, and initial voltage to be 120. (Scale as in 
question 21.) 

24. Plot the curve of electrical efficiency for the Wood 
dynamo (see data page 25). (Make ordinates 20% per inch, and 
abscissae 2 amperes per inch.) 

25. («) Form a resistance scale on the plot for the series 

dynamo in answer to question 22. (5) What was the resistance 

of the external circuit when the maximum current was flowing? 
(c) What was the total resistance of the circuit ? 

Jin answering this and following questions requiring plots, the student should 
use sheets of co-ordinate paper, size about 6 inches by 9 inches, the smallest 
divisions being .1 inch with lines every inch apart made heavy. The student 
may draw these neatly himself, or will be supplied by the school upon request. 





DIRECT CURRENT DYNAMOS. 


79 


26. (a) Wliat is the value of the maximum horse-power 
indicated on Fig 12 ? (ft) What was the resistance of the circuit 
at that point ? 

27. If the field coils of a shunt dynamo have a resistance of 
600 ohms and the voltage at the brushes is kept constant at 520 
volts, how much energy expressed in watts is lost in the field 
coils ? 

28. (a) A compound dynamo running at no load has an 

E. M. F. of 112 volts. At full load the E. M. F. at the brushes 
is 120 volts. For what percentage is the dynamo overcompounded ? 
(ft) If 50 horse-power is required to drive the armature of a 
dynamo which is delivering 38,000 watts to the external circuit, 
what is the commercial efficiency at this load? (e) If 35,000 
watts are delivered to the outside circuit of a dynamo which 
requires 39,000 watts to run it, what percentage of the energy is 
lost ? ( d ) If a dynamo that is delivering 40,000 watts has a 

commercial efficiency of 93%, how much energy expressed in 
horse-power is required to run the armature ? 

29. What becomes of the energy lost in a dynamo, that is, 
the energy that represents the difference between the input and 
the output of a dynamo ? 

30. A shunt-wound dynamo is to run at a constant E. M. F. 

of 180 volts. At this voltage and at no load the regulating rheo¬ 
stat which is in series with the shunt coils has all of its resistance 
in and the current flowing is 1.5 amperes. At full load a current 
of 1.8 amperes is necessary to magnetize the field coils in order 
that the full voltage should be maintained, (a) How many ohms 
resistance must be cut out of the rheostat? (ft) An ordinary 
compound dynamo with an initial E. M. F. of 500 volts is over- 
compounded for 5 % at full load. At full load the current in 
the shunt field coils is 5 amperes. On account of excessive loss 
of voltage in the distributing mains it is desired at full load to 
increase the E. M. F. at the brushes to 550 volts. In order to do 
this a current of 5.5 amperes must flow through the shunt field 
coils. How much resistance must be cut out of the regulating 
rheostat in series with the shunt field coils in order that the volt¬ 
age shall be raised to 550 ? Ans. 5 ohms. 








































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